Assume that the probability that a bomb dropped from an airplane will strike a certain target is 0.2. If 6 bombs are dropped, find the probability that
i. exactly 2 will strike the target.
ii. at least 2 will strike the target.
We have been given that, the probability that a bomb dropped from an airplane will strike a certain target is 0.2.
Also, that 6 bombs are dropped.
Let p be the probability that a bomb dropped from an airplane will strike a certain target.
Then, q is the probability that a bomb dropped from an airplane will not strike a certain target.
⇒ p = 0.2
We know that, p + q = 1
⇒ q = 1 – p
Let X be a random variable the represents the number of bombs that strike the target.
Then, the probability that r bombs strike the target out of n bombs is given by,
P (X = r) = nCrprqn-r
Where n = 6
Let us put the values of n, p, and q in the above equation.
…(A)
(i). We need to find the probability that exactly 2 will strike the target out of 6 bombs.
Probability is given by,
Probability = P (X = 2)
So, put r = 2 in equation (A).
⇒ Probability = 0.24576
∴, the probability that exactly 2 will strike the target out of 6 bombs is 0.24576.
(ii). We need to find the probability that at least 2 will strike the target out of 6 bombs.
Probability is given by,
Probability = P (X ≥ 2)
This can also be written as,
Probability = 1 – P (X < 2)
⇒ Probability = 1 – [P (X = 0) + P (X = 1)]
Let us find the value of P (X = 0).
For this, put r = 0 in equation (A).
Now, let us find the value of P (X = 1).
For this, put r = 0 in equation (A).
Now, putting all these values in 1 – [P (X = 0) + P (X = 1)].
⇒ Probability = 0.345
∴, the probability that at least 2 will strike the target out of 6 bombs is 0.345.