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## Mathematics

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**Session**Probability -2**Session Objectives**• Addition Theorem on Probability • Conditional Probability • Multiplication Theorem on Probability • Independent Events • Class Exercise**Addition Theorem For 2 Events**If A and B be are events in a sample space S. Then the probability of occurrence of at least one of the events A and B is given by**Addition Theorem For 3 Events**If A, B and C are three events in the sample space S. Then, If A, B and C are three mutually exclusive events, then**Event A or not A**The events ‘A’ and ‘not A’ are mutually exclusive events as each outcome of the experiment is either favourable to ‘A’ or ‘not A’. Therefore, the event ‘A or not A’ is a certain (or sure) event whose probability is 1.**Given two mutually exclusive events A and B such**and , find P(A or B). Example-1**Example-2**An integer is chosen at random from the first 200 positive integers. Find the probability that the integer is divisible by 6 or 8. Solution: Let S be sample space. Then, S = {1, 2, 3, …200}, n(S) = 200 Let A : event that the number is divisible by 6. A = {6, 12, 18 ... 198}, n(A) = 33 Let B : event that number is divisible by 8. B = {8, 16, 24 ... 200}, n(B) = 25**Solution Cont.**(A Ç B) : event that the number is divisible by 6 and 8. A Ç B = {24, 48, ... 192}, n(A Ç B) = 8 A B : event that the number is divisible either by 6 or 8.**Example-3**Find the probability of drawing a black king and probability of drawing a black card or a king from a well-shuffled pack of 52 cards. Solution: Let A and B be two events such that A : drawing a black card, B : drawing a king Total number of cards n(S) = 52 Number of black cards n(A) = 26 Number of kings n(B) = 4**Solution (Cont.)**and P(drawing a black card or a king)**Example-4**Find the probability of getting an even number on the first die or a total of 8 in a single through of two dice. (CBSE 2004) Solution: Let S be sample, then n(S) = 36 Let A and B be two events such that A = getting an even number on first die, B = getting a total of 8**Solution Cont.**A ={(2, 1), (2, 2), …, (2, 6), (4, 1), (4, 2), …, (4, 6), (6, 1), (6, 2), …, (6, 6)}, B = {(2, 6), (6, 2), (3, 5), (5, 3), (4, 4)}**Example-5**Two dice are tossed together. Find the probability that the sum of the numbers obtained on the two dice is neither a multiple of 3 nor a multiple of 4. (CBSE 1996) Solution: Let S be sample, then n(S) = 36 Let A and B be two events such that A = the sum of the numbers obtained on the two dice is a multiple of 3., B = the sum of the numbers obtained on the two dice is a multiple of 4**Solution (Cont.)**Then, A = {(1, 2), (1, 5), (2, 1), (2, 4), (3, 3), (3, 6), (4, 2), (4, 5), (5, 1), (5, 4), (6, 3), (6, 6)}, B = {(1, 3), (2, 2), (2, 6), (3, 1), (3, 5), (4, 4), (5, 3), (6, 2), (6, 6)}**In a random experiment, if A and B are two events, then the**probability of occurrence of event A when event B has already occurred and , is called the conditional probability and it is denoted by Conditional Probability**Multiplication Theorem on Probability**If A and B are two events associated with a random experiment, then**Solution: Given that P(A) = 0.5, P(B) = 0.6 and**Example-6 By addition theorem**Solution Cont.**By multiplication theorem**Example –7**A die is rolled twice and the sum of the numbers appearing on them is observed to be 7. What is the conditional probability that the number 2 has appeared at least once. (CBSE 1990, 2003) Solution: Let A and B be two events such that A = getting number 2 at least once B = getting 7 as the sum of the numbers on two dice A = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (1, 2), (3, 2), (4, 2), (5, 2), (6, 2)}**Solution (Cont.)**B = {(2, 5), (5, 2), (6, 1), (1, 6), (3, 4), (4, 3)}**Independent Events**Two events A and B are said to be independent if occurrence of one does not affect the occurrence of other; then**The probability of student A passing an examination is**• and the probability of student B passing is . Find the probability of • only A passing the examination • (ii) only one of them passing the examination Solution: (i) Let A : event of A passing the examination P(A') = 1 – P(A) Example –8**Let B : event of B passing the examination**P(B’) = 1 – P(B) Solution Cont. [As A and B' are independent events]**Solution Cont.**(ii) P(only one passing the examination) [As A and B' are independent events and A’ and B are also independent events]**Example –9**Three bags contain 7 white 7 red, 5 white 8 red, and 8 white 6 red balls respectively. One ball is drawn at random from each bag. Find the probability that the three balls drawn are of the same colour. Solution: W1 : White ball from bag I W2 : White ball from bag II W3 : White ball from bag III R1 : Red ball from bag I R2 : Red ball from bag II R3 : Red ball from bag III**Solution (Cont.)**P(Three balls are of same colour) [As W1, W2 and W3 are independent events and R1, R2 and R3 are also independent events]**Example –10**A speaks truth in 60% of the cases and B in 90% of the cases. In what percentage of the cases they are likely to contradict each other in stating the same fact? (CBSE 2003) Solution: Let A : event of A speaking the truth. B : event of B speaking the truth.**Solution (Cont.)**P(A and B contradicting each other) [As A and B' are independent of each other and A' and B are also independent of each other.**The probabilities of A, B and C solving a problem are**respectively. If the problem is attempted by all simultaneously, find the probability of exactly one of them solving it. (CBSE 2004) Example –11**Solution (Cont.)**Required probability [As A, B’ and C’ are independent events; A’, B and C’ are independent events; A’, B’ and C are also independent events]