Question

A magic shop sells a coin which is rather biased. It comes up heads 40% of the time. Suppose the shop owner puts on a demonstration his customers by flipping the coin 12 times. For each item below, write the letter which corresponds to the correct value. Some letters may not be used and some letters may be used more than once.

• Probability of exactly 6 heads?

• Probability of exactly 9 tails?

• Probability of more than than 2 heads?

• Probability of at least 5 heads?

• Probability of less than 3 heads?

• Mean/Expected value for the number of heads?

• Variance for the number of heads?

Choices: A. 0.012 B. 0.142 C. 0.158 D. 0.177 E. 0.775 F. 0.917 G. 1.70 H. 2.88 I. 4.80 J. None of the above.

Answer 1

p = 0.4

n = 12

This is a binomial distirbution.

P(X = x) = 12Cx * 0.4^x * (1 - 0.4)^12-x

a) P(X = 6) = 12C6 * 0.4⁶ * 0.6⁶ = 0.177 (choice -D)

b) P(9 tails) = P(3 heads) = P(X = 3) = 12C3 * 0.4³ * 0.6⁹ = 0.142 (choice -B)

c) P(X > 2) = 1 - (P(X = 0) + P(X = 1) + P(X = 2))

                  = 1 - (12C0 * 0.4⁰ * 0.6¹² + 12C1 * 0.4¹ * 0.6¹¹ + 12C2 * 0.4² * 0.6¹⁰ )

                  = 1 - 0.083

                  = 0.917 (choice - F)

d) P(X > 5) = 1 - (P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4))

                  = 1 - (12C0 * 0.4⁰ * 0.6¹² + 12C1 * 0.4¹ * 0.6¹¹ + 12C2 * 0.4² * 0.6¹⁰ + 12C3 * 0.4³ * 0.6⁹ + 12C4 * 0.4⁴ * 0.6⁸)

                  = 1 - 0.438

                  = 0.562 (choice - J)

e) P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 12C0 * 0.4⁰ * 0.6¹² + 12C1 * 0.4¹ * 0.6¹¹ + 12C2 * 0.4² * 0.6¹⁰ = 0.083 (choice - J)

f) Mean = n * p = 12 * 0.4 = 4.8 (Choice - I)

g) Variance = n * p * (1 - p) = 12 * 0.4 * 0.6 = 2.88 (choice - H)