Question

The probability that a ship safely reaches a port is $ \dfrac{1}{3} $ . The probability that out of 5 ships, at least 4 ships would arrive safely is
a) $ \dfrac{1}{{243}} $
b) $ \dfrac{{10}}{{243}} $
c) $ \dfrac{{11}}{{243}} $
d) $ \dfrac{{13}}{{243}} $

Answer 1

Hint: First we will find the probability of a ship that does not reach port safely. Then we will find probability that at least 4 arrivals are safe, for that reason we have to find p(4) and p(5) respectively. We will find $ ^5{C_4}{\left( {\dfrac{1}{3}} \right)^4}\left( {\dfrac{2}{3}} \right) $ for p(4) and $ ^5{C_5}{\left( {\dfrac{1}{3}} \right)^5} $ p(5). Then we will add both the terms to get the answer.
Formula used:
If a random experiment is done k- times and the probability that an event will occur is p and the probability of that event do not occur is q, then the probability that the event will occur exactly ‘n’ times is given by the formula,
 $ ^k{C_n}{p^n}{q^{k - n}} $
Where, $ ^k{C_n} = \dfrac{{k!}}{{n! \times (k - n)!}} $

Complete step-by-step answer:
Here we are given that the probability that a ship safely reaches a port is $ \dfrac{1}{3} $ .
So, to find the probability that it does not reach safely is basically $ 1 - $ the probability that it reaches safely.
Let s denotes safe reaching and s’ denotes not reaching safe,
 $ \Rightarrow P(s) = \dfrac{1}{3} $
 $ \Rightarrow P(s') = 1 - \dfrac{1}{3} = \dfrac{2}{3} $
Now we will find the probability that at least 4 arrives safely. We have a total 5 ships. So at least 4 arrives safely means 4 out of 5 arrives safely and 5 out of 5 arrives safely.
 P(at least 4 arrive safely) = P(4) + P(5)
So we will find out the probability of both the cases separately and find out the values and then we will add these numbers.
For 4 out of 5 case $ { = ^5}{C_4}{\left( {\dfrac{1}{3}} \right)^4}\dfrac{2}{3} $
 $ = \dfrac{{5!}}{{4! \times (5 - 4)!}}{\left( {\dfrac{1}{3}} \right)^4}\dfrac{2}{3} $
 $ = \dfrac{{5!}}{{4! \times 1!}} \times \dfrac{1}{{{3^4}}} \times \dfrac{2}{3} $
 $ = 5 \times \dfrac{1}{{{3^4}}} \times \dfrac{2}{3} $
 $ = 5 \times \dfrac{2}{{{3^5}}} $
 $ = \dfrac{{10}}{{{3^5}}} $
 $ = \dfrac{{10}}{{{3^5}}} $ --(1)
For 5 out of 5 case $ { = ^5}{C_5}{\left( {\dfrac{1}{3}} \right)^5} $
 $ = \dfrac{{5!}}{{5! \times (5 - 5)!}}{\left( {\dfrac{1}{3}} \right)^5} $
 $ = \dfrac{{5!}}{{5! \times 0!}}{\left( {\dfrac{1}{3}} \right)^5} $
 $ = 1 \times {\left( {\dfrac{1}{3}} \right)^5} $
 $ = {\left( {\dfrac{1}{3}} \right)^5} $ --(2)
Now adding (1) and (2) we get the probability of at least four ships reach safely,
 $ = \dfrac{{10}}{{{3^5}}} + \dfrac{1}{{{3^5}}} = \dfrac{{11}}{{{3^5}}} = \dfrac{{11}}{{243}} $
So the probability that at least four ships arrive safely is $ \dfrac{{11}}{{243}} $ .
So, the correct answer is “ $ \dfrac{{11}}{{243}} $ (OPTION C) ”.

Note: If we are provided with probability of a certain thing and we have to find the probability of another one then we have to subtract that number from 1 to get the probability of the same. The probability of any random experiment is always a positive term and can never exceed 1.