MATH SOLVE

3 months ago

Q:
# 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.

Accepted Solution

A:

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 = [tex]\frac{1}{2}
[/tex]Probability of not getting head, q = 1 - p = [tex]\frac{1}{2}
[/tex]Now, by Binomial distribution with:p = [tex]\frac{1}{2}
[/tex] = 0.5q = [tex]\frac{1}{2}
[/tex] = 0.5n = 6 P(X = r) = [tex]^{n}C_{r} p^{r} q^{n - r}[/tex]P(X = 4) = [tex]^{6}C_{4} p^{4} q^{2}[/tex]P(X = 4) = [tex]^{6}C_{4} \times 0.5^{4}\times 0.5^{2}[/tex]P(X = 4) = [tex]\frac{6!}{4!(6 - 4)!} \times 0.5^{4}\times 0.5^{2}[/tex]On solving the above eqn, we get:P(X = 4) = 0.2344Therefore, the probability of getting exactly 5 heads is 0.2344