During war one ship in ten was sunk in the average in making a certain voyage.what was the probability that at least three out of a country of six ships would arrive safely?
Answers
During war one ship in ten was sunk in the average in making a certain voyage. The probability that at least three out of a country of six ships would arrive safely is 0.998
In the question it is stated that during war, one ship in ten was sunk in average in making a certain voyage
Probability that the ship would sink = 1÷10 = 0.1 and probability that it would return (p) = 1 - 0.1 = 0.9
To find the probability that out of 6 ships, at least 3 would return, we have to use Binomial distribution i.e. for [P(X≥3) . It is also equal to 1 - [P(X<3)] i.e. for 1- {[P(X=0)], [P(X=1)] along with [P(X=2)].}
For [P(X=y)], representation is nCy (p)^y × (1-p)^(n-y).
In this problem, n=6, p=0.9, y=0,1,2.
∴P = 1 - [P(X<3)] = 1 - {[P(X=0)] + [P(X=1)] + [P(X=2)]}
[P(X=0)] = 6C0 (0.9)^0 × (0.1)^6 = 10^-6.
[P(X=1)] = 6C1 (0.9)^1 × (0.1)^5 = 5.4 × 10^-5.
[P(X=2)] = 6C2 (0.9)^2 × (0.1)^4 = 1.22 × 10^-3.
∴P = 1 - [P(X<3)] = 1 - {[P(X=0)] + [P(X=1)] + [P(X=2)]}
= 1 - (10^-6 + 5.4 × 10^-5 + 1.22 × 10^-3)
= 1 - (1.28 × 10^-3)
= 0.998
Thus, the probability that at least three out of a country of six ships would arrive safely is 0.998