Question

An experiment consists of tossing 7 fair​ (not weighted)​ coins, except one of the 7 coins has a head on both sides. Compute the probability of obtaining exactly 5 heads.

Answer 1

Answer:

Step-by-step explanation:

Out of the 7 coins, one is fake and has heads on both sides. So, fake coin on toss coin will always turns up to be head.

Therefore, to obtain  the probability of exactly 5 heads is:

probability of getting exactly 4 heads from the 6 fair coins

Probability of getting heads, p = $\frac{1}{2}$

Probability of not getting head, q = 1 - p = $\frac{1}{2}$

Now, by Binomial distribution with:

p = $\frac{1}{2}$ = 0.5

q = $\frac{1}{2}$ = 0.5

n = 6

P(X = r) = $^{n}C_{r} p^{r} q^{n - r}$

P(X = 4) = $^{6}C_{4} p^{4} q^{2}$

P(X = 4) = $^{6}C_{4} \times 0.5^{4}\times 0.5^{2}$

P(X = 4) = $\frac{6!}{4!(6 - 4)!} \times 0.5^{4}\times 0.5^{2}$

On solving the above eqn, we get:

P(X = 4) = 0.2344

Therefore, the probability of getting exactly 5 heads is 0.2344