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SYLLABUS: Random experiment – Outcomes – Sample space – Events – Occurrence of events – ‘not’, ‘and’ and ‘or’ events – Exhaustive events – Mutually exclusive events – Classical definition of probability – Axioms of probability – Probability of an event – Probability of ‘not’, ‘and’ and ‘or’ events – Conditional probability – Multiplication rule – Independent events – Simple problems (Combinatorial problems excluded) – Engineering applications (not for examinations).
The probability theory is an important branch of Mathematics. The word probability is used to denote the happening of certain event and likelihood of the occurrence of that event based on the past experiences.
5.1 PROBABILITY OF AN EVENT
Experiment: An experiment is defines as a process for which its result is well defined.
These are two types
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Deterministic experiment: An experiment whose outcomes can be predicted with certain, under identical conditions.
Random experiment: An experiment whose all possible outcomes are know, but it is not possible to predict the outcome.
Examples:
- A fair coin is “tossed”
- A die is “rolled”.
- Selecting a card from a pack of cards.
Outcome
The result of a random experiment is called an outcome.
Trial
Each performance of a random experiment is called a trial.
Sample space
The set of all possible outcomes of a random experiment is called a sample space. It is denoted by S.
Examples
- If a die is rolled, then the sample space S = {1,2,3,4,5,6}
- A coin is tossed, then the sample space S = {H,T}
- Two coins are tossed once. Then the sample space S = {HH,HT,TH,TT}
Event: Every non-empty subset of the sample space is an event. An event is the outcomes
of the random experiment
Notations
Let A and B be two events.
\[(i)\ A\ \cup\ B\ stands\ for\ the\ occurence\ of\ A\ or\ B\ or\ both\]
\[(ii)\ A\ \cap\ B\ stands\ for\ the\ simultaneous\ occurence\ of\ A\ and\ B\]
\[(iii)\ \bar{A}\ or\ A{\prime}\ or\ A^c\ stands\ for\ \ non-occurence\ of\ A\]
\[(ii)\ (A\ \cap\ \bar{B})\ stands\ for\ the\ occurence\ of\ only\ A\]
\[\color {purple} {Example\ 1\ .}\ A\ die\ is\ rolled\ once,\ \hspace{15cm}\]\[\color {red} {Find\ the\ probability\ of\ getting\ an\ odd\ number.}\ \hspace{5cm}\]
\[\hspace{5cm}\ October\ 2023\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\text{A standard die has 6 faces numbered from 1 to 6. No. of possible outcomes = 6}\]
\[\text{The odd numbers on a die are 1, 3, and 5. No. of favourable outcomes = 3}\]
\[\text{The probability P of rolling an odd number can be calculated as follows:}\]
\[P(odd\ number)\ =\ \frac{No.\ of\ favourable\ outcomes}{No.\ of\ possible\ outcomes}\ =\ \frac{3}{6}\ =\ \frac{1}{2}\]
\[\text{So, the probability of getting an odd number on a rolling die is:}\ \boxed{\frac{1}{2}}\]
\[\color {purple} {Example\ 2\ .}\ While\ a\ die\ is\ rolled\ once,\ \hspace{15cm}\]\[\color {red} {Find\ the\ probability\ of\ getting\ an\ even\ number}\ \hspace{5cm}\]
\[\hspace{5cm}\ October\ 2024\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\text{A standard die has 6 faces numbered from 1 to 6. No. of possible outcomes = 6}\]
\[\text{The even numbers on a die are 2, 4, and 6. No. of favourable outcomes = 3}\]
\[\text{The probability P of rolling an odd number can be calculated as follows:}\]
\[P(even\ number)\ =\ \frac{No.\ of\ favourable\ outcomes}{No.\ of\ possible\ outcomes}\ =\ \frac{3}{6}\ =\ \frac{1}{2}\]
\[\text{So, the probability of getting an odd number on a rolling die is:}\ \boxed{\frac{1}{2}}\]
\[\color {purple} {Example\ 3\ .}\ A\ die\ is\ rolled\ once,\ \hspace{15cm}\]\[\color {red} {Find\ the\ probability\ of\ getting\ a\ prime\ number.}\ \hspace{5cm}\]
\[\hspace{5cm}\ April\ 2024\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\text{A standard die has 6 faces numbered from 1 to 6. No. of possible outcomes = 6}\]
\[\text{The prime numbers on a die are 2, 3, and 5. No. of favourable outcomes = 3}\]
\[\text{A standard die has 6 faces numbered from 1 to 6. The prime numbers on a die are 2, 3, and 5.}\]
\[\text{The probability P of rolling a prime number can be calculated as follows:}\]
\[P(prime\ number)\ =\ \frac{Number\ of\ favourable\ outcomes}{Number\ of\ possible\ outcomes}\ =\ \frac{3}{6}\ =\ \frac{1}{2}\]
\[\text{So, the probability of getting a prime number on a rolling die is:}\ \boxed{\frac{1}{2}}\]
\[\color {purple} {Example\ 4\ .}\ \text{ An integer is chosen at random from the integers 1 to 10. }\ \hspace{10cm}\]\[\color {red} {Find\ the\ probability\ that\ it\ is\ an\ even\ number.}\ \hspace{5cm}\]
\[\hspace{5cm}\ October\ 2024\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\text{Given an integer is chosen at random from 1 to 10. No. of possible outcomes = 10}\]
\[\text{The even numbers from 1 to 10 are 2, 4, 6, 8, 10. No. of favourable outcomes = 5}\]
\[\text{The probability P of rolling a prime number can be calculated as follows:}\]
\[P(even\ number)\ =\ \frac{Number\ of\ favourable\ outcomes}{Number\ of\ possible\ outcomes}\ =\ \frac{5}{10}\ =\ \frac{1}{2}\]
\[\text{So, the probability of getting an even number from the integers 1 to 10 is:}\ \boxed{\frac{1}{2}}\]
\[\color {purple} {Example\ 5\ .}\ Two\ coins\ are\ tossed\ simultaneously\ \ \hspace{15cm}\]\[\color {red} {What\ is\ the\ probability\ of\ getting\ (i)\ exactly\ one\ head\ (ii)\ atleast\ one\ head\ (iii)\ atmost\ one\ head}\ \hspace{5cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[S\ =\ \text{{HH, HT, TH, TT}}\ \hspace{10cm}\]
\[n(S)\ =\ 4\ \hspace{15cm}\]
\[\text{Let A be the event of getting exactly one head}\ \hspace{10cm}\]
\[A\ =\ \text{{HT, TH}}\ \hspace{10cm}\]
\[n(A)\ =\ 2\ \hspace{15cm}\]
\[\text{Let B be the event of getting atleast one head}\ \hspace{10cm}\]
\[B\ =\ \text{{HH,HT, TH}}\ \hspace{10cm}\]
\[n(B)\ =\ 3\ \hspace{15cm}\]
\[\text{Let C be the event of getting atmost one head}\ \hspace{10cm}\]
\[C\ =\ \text{{TT,HT, TH}}\ \hspace{10cm}\]
\[n(C)\ =\ 3\ \hspace{15cm}\]
\[(i)\ P(A)\ =\ \frac{n(A)}{n(S)}\ =\ \frac{2}{4}\ =\ \frac{1}{2}\]
\[(ii)\ P(B)\ =\ \frac{n(B)}{n(S)}\ =\ \frac{3}{4}\]
\[(iii)\ P(C)\ =\ \frac{n(C)}{n(S)}\ =\ \frac{3}{4}\]
\[\color {purple} {Example\ 6\ .}\ Three\ coins\ are\ tossed\ simultaneously\ \color {red} {Find\ the\ probability\ of\ getting}\ \hspace{5cm}\]\[(i)\ exactly\ one\ head\ \hspace{13cm}\]\[(ii)\ exactly\ two\ heads\ \hspace{13cm}\]\[(iii)\ at least\ two\ heads\ \hspace{13cm}\]
\[\hspace{5cm}\ October\ 2023\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[S\ =\ \text{{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}}\ \hspace{10cm}\]
\[n(S)\ =\ 8\ \hspace{15cm}\]
\[\text{Let A be the event of getting exactly one head}\ \hspace{10cm}\]
\[A\ =\ \text{{HTT, THT, TTH}}\ \hspace{10cm}\]
\[n(A)\ =\ 3\ \hspace{15cm}\]
\[\text{Let B be the event of getting exactly two heads}\ \hspace{10cm}\]
\[B\ =\ \text{{HHT, HTH, THH}}\ \hspace{10cm}\]
\[n(B)\ =\ 3\ \hspace{15cm}\]
\[\text{Let C be the event of getting atleast two heads}\ \hspace{10cm}\]
\[C\ =\ \text{{HHT, HTH, THH, HHH}}\ \hspace{10cm}\]
\[n(C)\ =\ 4\ \hspace{15cm}\]
\[(i)\ P(A)\ =\ \frac{n(A)}{n(S)}\ =\ \frac{3}{8}\]
\[(ii)\ P(B)\ =\ \frac{n(B)}{n(S)}\ =\ \frac{3}{8}\]
\[(iii)\ P(C)\ =\ \frac{n(C)}{n(S)}\ =\ \frac{4}{8}\ =\ \frac{1}{2}\]
\[\color {purple} {Example\ 7\ .}\ Three\ coins\ are\ tossed\ simultaneously\ \color {red} {Find\ the\ probability\ of\ getting}\ \hspace{5cm}\]\[(i)\ at\ least\ one\ head\ \hspace{13cm}\]\[(ii)\ at\ most\ one\ head\ \hspace{13cm}\]\[(iii)\ exactly\ one\ head.\ \hspace{13cm}\]
\[\hspace{5cm}\ April\ 2024\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[S\ =\ \text{{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}}\ \hspace{10cm}\]
\[n(S)\ =\ 8\ \hspace{15cm}\]
\[\text{Let A be the event of getting at least one head}\ \hspace{10cm}\]
\[A\ =\ \text{{HTT, TTH, HHT, HTH, THT, THH, HHH}}\ \hspace{10cm}\]
\[n(A)\ =\ 7\ \hspace{15cm}\]
\[\text{Let B be the event of getting at most one head}\ \hspace{10cm}\]
\[B\ =\ \text{{TTT, HTT, THT, TTH}}\ \hspace{10cm}\]
\[n(B)\ =\ 4\ \hspace{15cm}\]
\[\text{Let C be the event of getting exactly one head}\ \hspace{10cm}\]
\[C\ =\ \text{{HTT, THT, TTH}}\ \hspace{10cm}\]
\[n(C)\ =\ 3\ \hspace{15cm}\]
\[(i)\ P(A)\ =\ \frac{n(A)}{n(S)}\ =\ \frac{7}{8}\]
\[(ii)\ P(B)\ =\ \frac{n(B)}{n(S)}\ =\ \frac{4}{8}\ =\ \frac{1}{2}\]
\[(iii)\ P(C)\ =\ \frac{n(C)}{n(S)}\ =\ \frac{3}{8}\]
\[\color {purple} {Example\ 8\ .}\ A\ card\ is\ picked\ randomly\ from\ a\ pack\ of\ 52\ cards.\ \hspace{8cm}\]\[\color {red} {Find\ the\ probability\ of\ getting\ a\ King}\ \hspace{10cm}\]
\[\hspace{5cm}\ October\ 2023\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
In a standard deck of 52 playing cards, there are 4 Kings (one from each suit: hearts, diamonds, clubs, and spades.
The probability PP of drawing a King from a deck of 52 cards can be calculated as follows:
\[P(King)\ =\ \frac{Number\ of\ Kings}{Total\ number\ of\ cards}\ =\ \frac{4}{52}\ =\ \frac{1}{13}\]
\[\text{So, the probability of picking a King from a standard deck of 52 cards is:}\ \boxed{\frac{1}{13}}\]
\[\color {purple} {Example\ 9\ .}\ A\ card\ is\ picked\ randomly\ from\ a\ pack\ of\ 52\ cards\ randomly.\ \color {red} {Find\ the\ probability}\ \hspace{8cm}\]\[of\ getting\ a\ queen\ card\ \hspace{10cm}\]
\[\hspace{5cm}\ April\ 2024\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
In a standard deck of 52 playing cards, there are 4 queens(one from each suit: hearts, diamonds, clubs, and spades).
The probability PP of drawing a queen from a deck of 52 cards can be calculated as follows:
\[P(King)\ =\ \frac{Number\ of\ queens}{Total\ number\ of\ cards}\ =\ \frac{4}{52}\ =\ \frac{1}{13}\]
\[\text{So, the probability of picking a queenfrom a standard deck of 52 cards is:}\ \boxed{\frac{1}{13}}\]
5.2 PROBABILITY OF ‘not’, ‘and’, ‘or’ EVENTS
Basic Results
\[1.\ P(\Phi)\ =\ 0\ \hspace{19cm}\]
\[2.\ If\ \bar{A}\ is\ the\ complimentary\ event\ of\ A,\ then\ P(\bar{A})\ =\ 1\ -\ P(A)\ \hspace{8cm}\]
\[3.\ P(A\ \cup\ B)\ =\ P(A)\ +\ P(B)\ – P(A\ \cap\ B)\ \hspace{12cm}\]
\[\color {purple} {Example\ 10\ .}\ If\ A\ and\ B\ are\ two\ events\ such\ that\ P(A)\ =\ 0.42,\ and\ P(B)\ =\ 0.48,\ \hspace{10cm}\]\[\color{red}{find\ P(\bar{A})\ and\ P(\bar{B})}\ \hspace{5cm}\]
\[\hspace{5cm}\ April\ 2024,\ October\ 2024\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[W.\ K.\ T\]
\[1.\ P(\bar{A})\ =\ 1\ -\ P(A)\]
\[2.\ P(\bar{B})\ =\ 1\ -\ P(B)\]
\[Given\ \hspace{15cm}\]
\[\bullet\ P(A)\ =\ 0.42\ \hspace{10cm}\]
\[\bullet\ P(B)\ =\ 0.48\ \hspace{10cm}\]
\[P(\bar{A})\ =\ 1\ -\ P(A)\ =\ 1\ -\ 0.42\ =\ 0.58\]
\[P(\bar{B})\ =\ 1\ -\ P(B)\ =\ 1\ -\ 0.48\ =\ 0.52\]
\[\color {purple} {Example\ 11\ .}\ If\ A\ and\ B\ are\ two\ events\ such\ that\ P(A)\ =\ 0.35,\ and\ P(B)\ =\ 0.62,\ \hspace{10cm}\]\[\color{red}{find\ P(\bar{A})\ and\ P(\bar{B})}\ \hspace{5cm}\]
\[\hspace{5cm}\ April\ 2025\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[W.\ K.\ T\]
\[1.\ P(\bar{A})\ =\ 1\ -\ P(A)\]
\[2.\ P(\bar{B})\ =\ 1\ -\ P(B)\]
\[Given\ \hspace{15cm}\]
\[\bullet\ P(A)\ =\ 0.35\ \hspace{10cm}\]
\[\bullet\ P(B)\ =\ 0.62\ \hspace{10cm}\]
\[P(\bar{A})\ =\ 1\ -\ P(A)\ =\ 1\ -\ 0.35\ =\ 0.65\]
\[P(\bar{B})\ =\ 1\ -\ P(B)\ =\ 1\ -\ 0.62\ =\ 0.38\]
\[\color {purple} {Example\ 12\ .}\ If\ P(A)\ =\ 0.15,\ P(B)\ =\ 0.25,\ and\ P(A\ \cap\ B)\ =\ 0.10,\ \hspace{10cm}\]\[\color {red} {find\ the\ value\ of\ P(A\ \cup\ B)}\ \hspace{10cm}\]
\[\hspace{5cm}\ Supp\ June\ 2025\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\text{Given P(A) = 0.15, P(B) = 0.25 and}\ P(A\ \cap\ B)\ =\ 0.10\]
\[We\ know\ that\]
\[P(A\ \cup\ B\ =\ P(A)\ +\ P(B)\ – P(A\ \cap\ B)\]
\[\hspace{2cm}\ =\ 0.15\ +\ 0.25\ -\ 0.10\]
\[\hspace{2cm}\ =\ 0.40\ -\ 0.10\]
\[\boxed{P(A\ \cup\ B)\ =\ 0.3}\]
\[\color {purple} {Example\ 13\ .}\ If\ P(A)\ =\ 0.5,\ P(B)\ =\ 0.3,\ and\ A\ \cap\ B\ is\ empty,\ \hspace{10cm}\]\[\color {red} {find\ P(A\ \cup\ B)}\ \hspace{10cm}\]
\[\hspace{5cm}\ October\ 2023\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\text{Given P(A) = 0.5, P(B) = 0.3 and} A \cap\ B\ is\ empty\ (\text{ which means events A and B are mutually exclusive),}\]
\[we\ can\ find\ P(A\ \cup\ B)\ \text{using the formula for the union of two mutually exclusive events:}\]
\[P(A\ \cup\ B\ =\ P(A)\ +\ P(B)\]
\[\hspace{2cm}\ =\ 0.5\ +\ 0.3\]
\[\boxed{P(A\ \cup\ B)\ =\ 0.8}\]
\[\color {purple} {Example\ 14\ .}\ Two\ dice\ are\ thrown\ simultaneously\ \color {red} {Find\ the\ probability}\ \hspace{10cm}\\\ of\ getting\ a\ sum\ 5\ or\ same\ number\ on\ both\ dice\ \hspace{5cm}\]
\[\hspace{5cm}\ April\ 2025\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[When\ two\ dice\ are\ thrown\ simultaneously,\ each\ die\ has\ 6\ faces,\ so\ there\ are\ a\ \hspace{2cm}\\\ total\ of\ 6\ \times\ 6\ =\ 36\ possible\ outcomes\ \hspace{12cm}\]
\[1.\ \bf{\color{green}{probability\ of\ getting\ a\ sum\ of\ 6}}\ \hspace{10cm}\]
\[\text{The possible outcomes where the sum of the two dice is 5 are:}\ \hspace{5cm}\]
\[\bullet\ (1,\ 4)\ \hspace{15cm}\]
\[\bullet\ (2,\ 3)\ \hspace{15cm}\]
\[\bullet\ (3,\ 2)\ \hspace{15cm}\]
\[\bullet\ (4,\ 1)\ \hspace{15cm}\]
\[\text{There are 4 such outcomes.}\ \hspace{10cm}\]
\[2.\ \bf{\color{green}{probability\ of\ getting\ the\ same\ number\ on\ both\ dice}}\ \hspace{7cm}\]
\[\text{The possible outcomes where the numbers on both dice are the same are:}\ \hspace{5cm}\]
\[\bullet\ (1,\ 1)\ \hspace{15cm}\]
\[\bullet\ (2,\ 2)\ \hspace{15cm}\]
\[\bullet\ (3,\ 3)\ \hspace{15cm}\]
\[\bullet\ (4,\ 4)\ \hspace{15cm}\]
\[\bullet\ (5,\ 5)\ \hspace{15cm}\]
\[\bullet\ (6,\ 6)\ \hspace{15cm}\]
\[\text{There are 6 such outcomes.}\ \hspace{10cm}\]
\[3.\ \bf{\color{green}{probability\ of\ either\ getting\ a\ \ sum\ of\ 5\ or\ the\ same\ number\ on\ both\ dice}}\ \hspace{2cm}\]
\[A\ denotes\ the\ event\ of\ getting\ a\ sum\ of\ 5\ \hspace{10cm}\]
\[B\ denotes\ the\ event\ of\ getting\ the\ same\ number\ on\ both\ dice\ \hspace{5cm}\]
\[we\ needed\ to\ find\ P(A\ \cup\ B) =\ ?\]
\[P(A\ \cup\ B)\ =\ P(A)\ +\ P(B)\ -\ P(A\ \cap\ B)\ \hspace{5cm}\]
\[P(A)\ =\ \frac{4}{36}\ \hspace{10cm}\]
\[P(B)\ =\ \frac{6}{36}\ \hspace{10cm}\]
\[P(P(A\ \cap\ B))\ =\ 0\ \hspace{10cm}\]
\[P(A\ \cup\ B)\ =\ \frac{4}{36}\ +\ \frac{6}{36}\ -\ 0\ \hspace{5cm}\]
\[=\ \frac{10}{36}\ =\ \frac{5}{18}\ \hspace{3cm}\]
\[\text{So, the probability of getting a sum of 5 or the same number on both dice is:}\ =\ \frac{5}{18}\ \hspace{1cm}\]
\[\color {purple} {Example\ 15\ .}\ Two\ dice\ are\ thrown\ simultaneously\ \color {red} {Find\ the\ probability}\ \hspace{10cm}\\\ of\ getting\ a\ sum\ 6\ or\ same\ number\ on\ both\ dice\ \hspace{5cm}\]
\[\hspace{5cm}\ October\ 2023\ Supp(June)\ 25\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[When\ two\ dice\ are\ thrown\ simultaneously,\ each\ die\ has\ 6\ faces,\ so\ there\ are\ a\ \hspace{2cm}\\\ total\ of\ 6\ \times\ 6\ =\ 36\ possible\ outcomes\ \hspace{12cm}\]
\[1.\ \bf{\color{green}{probability\ of\ getting\ a\ sum\ of\ 6}}\ \hspace{10cm}\]
\[\text{The possible outcomes where the sum of the two dice is 6 are:}\ \hspace{5cm}\]
\[\bullet\ (1,\ 5)\ \hspace{15cm}\]
\[\bullet\ (2,\ 4)\ \hspace{15cm}\]
\[\bullet\ (3,\ 3)\ \hspace{15cm}\]
\[\bullet\ (4,\ 2)\ \hspace{15cm}\]
\[\bullet\ (5,\ 1)\ \hspace{15cm}\]
\[\text{There are 5 such outcomes.}\ \hspace{10cm}\]
\[2.\ \bf{\color{green}{probability\ of\ getting\ the\ same\ number\ on\ both\ dice}}\ \hspace{7cm}\]
\[\text{The possible outcomes where the numbers on both dice are the same are:}\ \hspace{5cm}\]
\[\bullet\ (1,\ 1)\ \hspace{15cm}\]
\[\bullet\ (2,\ 2)\ \hspace{15cm}\]
\[\bullet\ (3,\ 3)\ \hspace{15cm}\]
\[\bullet\ (4,\ 4)\ \hspace{15cm}\]
\[\bullet\ (5,\ 5)\ \hspace{15cm}\]
\[\bullet\ (6,\ 6)\ \hspace{15cm}\]
\[\text{There are 6 such outcomes.}\ \hspace{10cm}\]
\[3.\ \bf{\color{green}{probability\ of\ either\ getting\ a\ \ sum\ of\ 6\ or\ the\ same\ number\ on\ both\ dice}}\ \hspace{2cm}\]
\[A\ denotes\ the\ event\ of\ getting\ a\ sum\ of\ 6\ \hspace{10cm}\]
\[B\ denotes\ the\ event\ of\ getting\ the\ same\ number\ on\ both\ dice\ \hspace{5cm}\]
\[we\ needed\ to\ find\ P(A\ \cup\ B) =\ ?\]
\[P(A\ \cup\ B)\ =\ P(A)\ +\ P(B)\ -\ P(A\ \cap\ B)\ \hspace{5cm}\]
\[P(A)\ =\ \frac{5}{36}\ \hspace{10cm}\]
\[P(B)\ =\ \frac{6}{36}\ \hspace{10cm}\]
\[P(P(A\ \cap\ B))\ =\ \frac{1}{36}\ \hspace{10cm}\]
\[P(A\ \cup\ B)\ =\ \frac{5}{36}\ +\ \frac{6}{36}\ -\ \frac{1}{36}\ \hspace{5cm}\]
\[=\ \frac{10}{36}\ =\ \frac{5}{18}\ \hspace{3cm}\]
\[\text{So, the probability of getting a sum of 6 or the same number on both dice is:}\ =\ \frac{5}{18}\ \hspace{1cm}\]
\[\color {purple} {Example\ 16\ .}\ Two\ dice\ are\ rolled\ together.\ \color {red} {Find\ the\ probability}\ \hspace{10cm}\\\ for\ getting\ the\ sum\ of\ the\ numbers\ on\ the\ faces\ are\ 10\ and\ 7\ \hspace{5cm}\]
\[\hspace{5cm}\ October\ 2024\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[When\ two\ dice\ are\ thrown\ simultaneously,\ each\ die\ has\ 6\ faces,\ so\ there\ are\ a\ \hspace{2cm}\\\ total\ of\ 6\ \times\ 6\ =\ 36\ possible\ outcomes\ \hspace{12cm}\]
\[1.\ \bf{\color{green}{probability\ of\ getting\ a\ sum\ of\ 10}}\ \hspace{10cm}\]
\[\text{The possible outcomes where the sum of the two dice is 6 are:}\ \hspace{5cm}\]
\[\bullet\ (4,\ 6)\ \hspace{15cm}\]
\[\bullet\ (5,\ 5)\ \hspace{15cm}\]
\[\bullet\ (6,\ 4)\ \hspace{15cm}\]
\[\text{There are 3 such outcomes.}\ \hspace{10cm}\]
\[A\ denotes\ the\ event\ of\ getting\ a\ sum\ of\ 10\ \hspace{10cm}\]
\[P(A)\ =\ \frac{3}{36}\ \hspace{10cm}\]
\[2.\ \bf{\color{green}{probability\ of\ getting\ a\ sum\ of\ 7}\ \hspace{7cm}\]
\[\text{The possible outcomes where the numbers on both dice are the same are:}\ \hspace{5cm}\]
\[\bullet\ (1,\ 6)\ \hspace{15cm}\]
\[\bullet\ (2,\ 5)\ \hspace{15cm}\]
\[\bullet\ (3,\ 4)\ \hspace{15cm}\]
\[\bullet\ (4,\ 3)\ \hspace{15cm}\]
\[\bullet\ (5,\ 2)\ \hspace{15cm}\]
\[\bullet\ (6,\ 1)\ \hspace{15cm}\]
\[\text{There are 6 such outcomes.}\ \hspace{10cm}\]
\[B\ denotes\ the\ event\ of\ getting\ a\ sum\ of\ 10\ \hspace{10cm}\]
\[P(B)\ =\ \frac{6}{36}\ \hspace{10cm}\]
\[3.\ \bf{\color{green}{probability\ of\ sum\ of\ the\ numbers\ on\ on\ the\ faces\ are\ 10\ and\ 7}}\ \hspace{2cm}\]
\[we\ needed\ to\ find\ P(A\ \cup\ B) =\ ?\]
\[P(A\ \cup\ B)\ =\ P(A)\ +\ P(B)\ -\ P(A\ \cap\ B)\ \hspace{5cm}\]
\[P(A)\ =\ \frac{3}{36}\ \hspace{10cm}\]
\[P(B)\ =\ \frac{6}{36}\ \hspace{10cm}\]
\[P(P(A\ \cap\ B))\ =\ 0\ \hspace{10cm}\]
\[P(A\ \cup\ B)\ =\ \frac{3}{36}\ +\ \frac{6}{36}\ -\ 0\ \hspace{5cm}\]
\[=\ \frac{9}{36}\ =\ \frac{1}{4}\ \hspace{3cm}\]
\[\text{So, the probability of getting a sum of 10 and 7 on both dice is:}\ =\ \frac{1}{4}\ \hspace{1cm}\]
\[\color {purple} {Example\ 17\ .}\ A\ card\ is\ selected\ at\ random\ from\ a\ pack\ of\ 52\ cards.\ \color{red}{Find\ the}\ \hspace{8cm}\]\[\color{red}{probability\ that\ the\ card\ is\ either\ a\ black\ card\ or\ a\ card\ with\ number\ 6.}\ \hspace{2cm}\]
\[\hspace{5cm}\ April\ 2024,\ October\ 2024\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\text{To find the probability of drawing either a black card or a card with the number 6 from a standard deck of 52 cards,}\\ \text{we need to consider both events and their possible overlap.}\ \hspace{10cm}\]
\[1.\ \text{Total number of cards in a deck: 52}\ \hspace{5cm}\]
\[\hspace{3cm}\ 2.\ \text{Number of black cards: There are 26 black cards in the deck (13 spades +}\\ \text{13 clubs}).\ \hspace{10cm}\]
\[\hspace{3cm}\ 3.\ \text{Number of cards with the number 6: There are 4 cards with the number 6}\\ \text{(one in each suit: hearts, diamonds, clubs, and spades}).\ \hspace{2cm}\]
\[A\ denotes\ the\ event\ of\ drawing\ a\ black\ card\ \hspace{10cm}\]
\[B\ denotes\ the\ event\ of\ drawing\ a\ card\ with\ on\ number\ 6\ \hspace{5cm}\]
\[we\ needed\ to\ find\ P(A\ \cup\ B) =\ ?\]
\[P(A\ \cup\ B)\ =\ P(A)\ +\ P(B)\ -\ P(A\ \cap\ B)\ \hspace{5cm}\]
\[P(A)\ =\ \frac{26}{52}\ \hspace{10cm}\]
\[P(B)\ =\ \frac{4}{52}\ \hspace{10cm}\]
\[P(P(A\ \cap\ B))\ =\ \frac{2}{52}\ (2\ such\ cards\ with \ black\ card\ and\ number\ 6\ from\ spades\ and\ clubs)\]
\[P(A\ \cup\ B)\ =\ \frac{26}{52}\ +\ \frac{4}{52}\ -\ \frac{2}{52}\ \hspace{5cm}\]
\[=\ \frac{28}{52}\ =\ \frac{7}{13}\ \hspace{3cm}\]
\[\text{So, the probability of drawing either a black card or a card with the}\\ \text{number 6 from a standard deck of 52 cards is:} \frac{7}{13}\ \hspace{5cm}\]
5.3 CONDITIONAL PROBABILITY
\[\text{The probability of occurrence of A when B has occurred is called the conditional probability}\]
\[A\ given\ B,\ denoted\ as\ P(A/B),\ is\ defied\ by\ P(A/B)\ =\ \frac{P(A\ \cap\ B}{P(B)},\ provided\ P(B)\ \neq\ 0.\ Similarly,\]
\[P(B/A)\ =\ \frac{P(A\ \cap\ B}{P(A)},\ provided\ P(A)\ \neq\ 0.\]
Independent events
Two events are said to be independent events if the occurrence of any one event does not affect the probability of occurrence of another event. Two events A and B are said to be independent if and only if P(A ∩ B) = P(A)P(B).
\[\color {purple} {Example\ 18\ .}\ \color{red}{Find\ P(A/B)},\ If\ P(B)\ =\ 0.5,\ and\ P(A\ \cap\ B)\ =\ 0.2,\ \hspace{10cm}\]
\[\hspace{5cm}\ April\ 2024\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[To\ find\ the\ conditional\ probability\ P(A/B),\ \text{We use the definitions of conditional probability.}\]
\[P(A/B)\ =\ \frac{P(A\ \cap\ B)}{P(B)}\]
\[Given\ \hspace{15cm}\]
\[\bullet\ P(B)\ =\ 0.5\ \hspace{10cm}\]
\[\bullet\ P(A\ \cap\ B)\ =\ \ 0.2\ \hspace{10cm}\]
\[P(A/B)\ =\ \frac{P(A\ \cap\ B)}{P(B)}\ =\ \frac{0.2} {0.5}\ =\ \frac{2}{5}\ =\ 0.4\]
\[\color {purple} {Example\ 19\ .}\ If\ P(A)\ =\ \frac{1}{3},\ P(B)\ =\ \frac{3}{4},\ P(A\ \cap\ B)\ =\ \frac{1}{6},\ \hspace{10cm}\]\[\color{red}{find\ P(A/B)\ and\ P(B/A)}\ \hspace{5cm}\]
\[\hspace{5cm}\ October\ 2023\ April\ 25\ Supp\ June\ 25\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[To\ find\ the\ conditional\ probabilities\ P(A/B)\ and\ P(B/A),\ \text{We use the definitions of conditional probability.}\]
\[1.\ P(A/B)\ =\ \frac{P(A\ \cap\ B)}{P(B)}\]
\[2.\ P(B/A)\ =\ \frac{P(A\ \cap\ B)}{P(A)}\]
\[Given\ \hspace{15cm}\]
\[\bullet\ P(A)\ =\ \frac{1}{3}\ \hspace{10cm}\]
\[\bullet\ P(B)\ =\ \frac{3}{4}\ \hspace{10cm}\]
\[\bullet\ P(A\ \cap\ B)\ =\ \frac{1}{6}\ \hspace{10cm}\]
\[P(A/B)\ =\ \frac{P(A\ \cap\ B)}{P(B)}\ =\ \frac{\frac{1}{6}} {\frac{3}{4}}\ =\ \frac{1}{6}\ \times\ \frac{4}{3}\ =\ \frac{2}{9}\]
\[P(B/A)\ =\ \frac{P(A\ \cap\ B)}{P(A)}\ =\ \frac{\frac{1}{6}} {\frac{1}{3}}\ =\ \frac{1}{6}\ \times\ \frac{3}{1}\ =\ \frac{1}{2}\]
\[\color {purple} {Example\ 20\ .}\ If\ A\ and\ B\ are\ two\ independent\ events\ such\ that\ P(A)\ =\ 0.4\ and\ P(A\ \cup\ B)\ =\ 0.9.\ \hspace{10cm}\]\[\color {red} {Find\ P(B)}\ \hspace{5cm}\]
\[\hspace{5cm}\ October\ 2024\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[Given\ \hspace{15cm}\]
\[\bullet\ P(A)\ =\ 0.4\ \hspace{10cm}\]
\[\bullet\ P(A\ \cup\ B)\ =\ 0.9\ \hspace{10cm}\]
\[\bullet\ P(A\ \cap\ B)\ =\ P(A)\ P(B)\ (A\ and\ B\ are\ independent\ events)\ \hspace{5cm}\]
\[P(A\ \cup\ B)\ =\ P(A)\ +\ P(B)\ -\ P(A\ \cap\ B)\ \hspace{5cm}\]
\[0.9\ =\ 0.4\ +\ P(B)\ -\ P(A)\ P(B)\ \ \hspace{5cm}\]
\[0.9\ -\ 0.4\ =\ P(B) (1\ -\ P(A))\ \hspace{5cm}\]
\[0.5\ =\ P(B) (1\ -\ 0.4)\ \hspace{5cm}\]
\[\boxed{P(B)\ =\ 5/6}\]
\[\color {purple} {Example\ 21\ .}\ A\ problem\ in\ statistics\ is\ given\ to\ two\ students\ A\ and\ B.\ The\ \hspace{10cm}\\\ probability\ of\ A\ solves\ the\ problem\ is\ \frac{1}{2}\ and\ that\ of\ B\ solves\ the\ \hspace{8cm}\\\ problem\ is\ \frac{2}{3}.\ If\ the\ students\ solve\ the\ problems\ independently.\ \hspace{8cm}\\\ \color{red}{find\ the\ probability\ that\ the\ problem\ is\ solved.}\ \hspace{12cm}\]
\[\hspace{5cm}\ October\ 2023\ April\ 25\ Supp\ June\ 25\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[Given,\ P(A)\ =\ \frac{1}{2}\ and\ P(B)\ =\ \frac{2}{3}\ \hspace{10cm}\\\ \text{Probability that the problem is solved = Probability that A solves the problem or B solves the problem}\]
\[=\ P(A\ \cup\ B)\ \hspace{5cm}\]
\[=\ P(A)\ +\ P(B)\ -\ (P(A\ \cap\ B)\ \hspace{2cm}\]
\[=\ P(A)\ +\ P(B)\ -\ P(A)\ \cdot \ P(B)\ Since A\ and\ B\ are\ independent\ \hspace{2cm}\]
\[=\ \frac{1}{2}\ +\ \frac{2}{3}\ -\ \frac{1}{2}\ \cdot \ \frac{2}{3}\ \hspace{2cm}\]
\[=\ \frac{1}{2}\ +\ \frac{2}{3}\ -\ \frac{1}{3}\ \hspace{2cm}\]
\[=\ \frac{3\ +4\ -\ 2}{6}\ \hspace{2cm}\]
\[=\ \frac{5}{6}\ \hspace{2cm}\]
\[\text{Thus, the probability that the problem is solved by either student A, student B, or both is:}\ =\ \frac{5}{6}\]
\[\color {purple} {Example\ 22\ .}\ A\ problem\ in\ statistics\ is\ given\ to\ two\ students\ A\ and\ B.\ The\ \hspace{10cm}\\\ probability\ of\ A\ solves\ the\ problem\ is\ \frac{1}{4}\ and\ that\ of\ B\ solves\ the\ \hspace{8cm}\\\ problem\ is\ \frac{2}{5}.\ If\ they\ solve\ the\ problem\ independently,\ \hspace{8cm}\\\ \color{red}{find\ the\ probability\ that\ the\ problem\ is\ solved.}\ \hspace{12cm}\]
\[\hspace{5cm}\ April\ 2024,\ October\ 2024\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[Given,\ P(A)\ =\ \frac{1}{4}\ and\ P(B)\ =\ \frac{2}{5}\ \hspace{10cm}\\\ \text{Probability that the problem is solved = Probability that A solves the problem or B solves the problem}\]
\[=\ P(A\ \cup\ B)\ \hspace{5cm}\]
\[=\ P(A)\ +\ P(B)\ -\ (P(A\ \cap\ B)\ \hspace{2cm}\]
\[=\ P(A)\ +\ P(B)\ -\ P(A)\ \cdot \ P(B)\ Since A\ and\ B\ are\ independent\ \hspace{2cm}\]
\[=\ \frac{1}{4}\ +\ \frac{2}{5
}\ -\ \frac{1}{4}\ \cdot \ \frac{2}{5}\ \hspace{2cm}\]
\[=\ \frac{1}{4}\ +\ \frac{2}{5}\ -\ \frac{1}{10}\ \hspace{2cm}\]
\[=\ \frac{5\ +\ 8\ -\ 2}{20}\ \hspace{2cm}\]
\[=\ \frac{11}{20}\ \hspace{2cm}\]
\[\text{Thus, the probability that the problem is solved by either student A, student B, or both is:}\ =\ \frac{11}{20}\]
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