Section 1.3 Conditional Probability 25 However, we can also find this probabilit

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Section 1.3 Conditional Probability 25 However, we can also find this probability by using combinations as follows: (3)(2) C) (2) - (10) (9) 15 (1)(2) Thus, we obtain the same answer, as we should, provided that our reasoning is consistent with the underlying assumptions. Example 1.3-7 From an ordinary deck of playing cards, cards are to be drawn successively at random and without replacement. The probability that the third spade appears on the sixth draw is computed as follows: Let A be the event of two spades in the first five cards drawn, and let B be the event of a spade on the sixth draw. Thus, the probability that we wish to compute is P(An B). It is reasonable to take /13)/39 ( 2 / 3) = 0.274 and P(B|A) = A = 0.234. P(A) = - The desired probability, P(ANB), is the product of those numbers: P(AN B) = (0.274)(0.234) = 0.064. nple -8 Continuing with Example 1.3-4, in which a pair of four-sided dice is rolled, the prob ability of rolling a sum of 3 on the first roll and then, continuing the sequence o rolls, rolling a sum of 3 before rolling a sum of 5 is 2 2 4 1 16 696 = 24 The multiplication rule can be extended to three or more events. In the ca at the multiplication rule for two events, we have

Explanation / Answer

Let A be the event that 3 heart appears in first 6 draws

Let B be the event of heart appears on 7th draw

Thus, we need to Compute P(A B)

3 hearts can be chosen from 13 in 13C3 ways. Remaining 3 can be chosen from 39C3. Total 6 out of 52 are chosen

Thus, P(A) = (13C3)(39C3) / (52C6) = 286*9139/20358520

= 0.128

P(B/A) = 10/46 (as 10 hearts are left after selecting 3 and 46 cards are left after picking 6)

Thus, P(A B) = P(A)*P(B/A)

= 0.128 * 10/46

= 0.028

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