GMIHIR wrote:George is aiming his arrows at the target shown by his mentor.If probability of hitting the target is 0.7, what is the approximate probability that he will hit the target only three times in his first five attempts?
A 0.1

B 0.3

C 0.85

D 0.5

E 0.65

Between C and E, E with a wild guess!

Is the answer B?
as Hit = .7 & miss = .3
Binomial probability distribution = 5C3 (.7)^3 * (.3)^2
=.3087

GMIHIR wrote:George is aiming his arrows at the target shown by his mentor.If probability of hitting the target is 0.7, what is the approximate probability that he will hit the target only three times in his first five attempts?
A 0.1

B 0.3

C 0.85

D 0.5

E 0.65

We need to determine the probability that George hits the target 3 times and misses twice, in 5 attempts.

Since the probability of a hit is 0.7, the probability of a miss is 0.3. Thus, we need to determine:

P(H-H-H-M-M) = 0.7 x 0.7 x 0.7 x 0.3 x 0.3 ≈ 0.03

However, we have to account for the number of ways we can arrange H-H-H-M-M, which is a combination problem because order doesn't matter.

5C3 = 5!/[3!(5-3)!] = 5!/(3! x 2!) = (5 x 4)/2 = 10 ways

Thus, P(3H and 2M) ≈ 0.03 x 10 = 0.3.

Answer: B