draw nine cards from a standard deck of cards. What is the probability that a. four or more will be red, b. exactly two or three will be red, c. two or fewer will be red, and d. exactly five will be red?
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The probabilities are governed by binomial distribution: (p+(1-p))^9=p^9+9p^8(1-p)+(9*8/1*2*)p^7(1-p)^2+...+(1-p)^9. The coefficients are given by the series 1 9 36 84 126 126 84 36 9 1 (Pascal's triangle) and p, the probability of picking a red card is 1/2, so 1-p, the probability of picking a black card, is also 1/2. The first term is the probability of picking 9 reds (or, in this case, 9 blacks); the second term is the probability of picking exactly 2 reds; and so on. We add together probabilities when the question implies OR and we multiply when the question implies AND.

a.

Four or more being red is the same as 1-(3 or fewer being black)=1-(1+9+36)*(1/2)^9=1-46/512=91.02%.

b.

Exactly 2 being red is the third term: 36*(1/2)^9; and exactly 3 is the fourth term 84*(1/2)^9. The probability of EITHER of these is the sum of these probabilities: 120*(1/2)^9=0.2344 or 23.44%.

c.

2 or fewer means 1 or 2, the sum of the first two terms: 10*(1/2)^9=1.95%.

d.

Exactly 5 is 126/512=63/256=24.61%.

 

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