Question

Consider rolling a fair die twice and tossing a fair coin eleven times. Assume that all the tosses and rolls are independent. The chance that the total number of heads in all the coin tosses equals 10 is (Q3)

Answer 1

Answer:

0.005371

Step-by-step explanation:

Given that a fair coin is tossed 11 times.

Since coin is fair probability for head p = 0.5 and q = prob for tail =0.5

Each trial is independent of the other. Hence X no of heads is binomial with $n = 11 and p =0.5$

Prob for getting 10 heads out of 11 tosses

=P(X=10)

=$11C10 (0.5)^{10} (0.5)\\= 0.005371$