Question

A dart is to be thrown at a target. The probability the dart will hit the target (yes or no) on a single attempt is 0.20. Each throw is independent of the other throws. Let X be the number of attempts before the target is hit.
a. Now say we want to hit the target twice. What is the probability that it will take less than or equal to 4 throws to hit the target on both successful target hits?

Answer 1

Answer:

0.04

Step-by-step explanation:

Given that a dart is to be thrown at a target. The probability the dart will hit the target (yes or no) on a single attempt is 0.20

Each throw is independent of the other throws.

Let x be the no of times the target is hit in the trial of 4 throws

Since each outcome is independent of the other X is binomial with n =4 and p = 0.20

Required probability

=  the probability that it will take less than or equal to 4 throws to hit the target on both successful target hits

=P(X=2 when n =2) + P(x=2 when x =3) + P(x=2 when x =4)

= $(0.2)^2 + 3C2 (0.2)^2(0.98)+4C2 (0.2)^2 (0.98^2)\\=0.04$