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Rework problem 11 from section 4.1 of your text involving the flipping of a weighted coin. Assume that the coin is weighted so that a head is 7 times as likely as a tail. The coin is flipped 8 times. What is the probability that both heads and tails occur?

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Answer:

0.2637

Explanation:

Probability of a tail turning up = x

Probability of an head turning up = 7x

x + 7x = 1

8x = 1

x = (1/8) = 0.125 = P(tail)

7x = (7/8) = 0.875 = P(head)

If the coin is flipped 8 times, the different combo for HEADS and TAILS include

2H, 6T

3H, 5T

4H, 4T

5H, 3T

6H, 2T

(Note that HEADS indicate a minimum of 2 and TAILS indicate a minimum of 2 too)

The probability of each of them can be found Using Binomial distribution

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = 8

x = Number of successes required = 2,3,4,5,6

p = probability of success = probabilty of a head turning up = 0.875

q = probability of failure = probability of a tail turning up = 0.125

when x = 2

P(2H, 6T) = 0.00008177757

when x = 3

P(3H, 5T) = 0.001144886

when x = 4

P(4H, 4T) = 0.010017753

when x = 5

P(5H, 3T) = 0.056099415

when x = 6

P(6H, 2T) = 0.19634795189

Total probability = sum of all of these = 0.2636917836 = 0.2637 to 4 d.p

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