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Question: The probability that a ship safely reaches a port is \( \dfrac{1}{3} \) . The probability that out o...

The probability that a ship safely reaches a port is 13\dfrac{1}{3} . The probability that out of 5 ships, at least 4 ships would arrive safely is
a) 1243\dfrac{1}{{243}}
b) 10243\dfrac{{10}}{{243}}
c) 11243\dfrac{{11}}{{243}}
d) 13243\dfrac{{13}}{{243}}

Explanation

Solution

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 5C4(13)4(23)^5{C_4}{\left( {\dfrac{1}{3}} \right)^4}\left( {\dfrac{2}{3}} \right) for p(4) and 5C5(13)5^5{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,
kCnpnqkn^k{C_n}{p^n}{q^{k - n}}
Where, kCn=k!n!×(kn)!^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 13\dfrac{1}{3} .
So, to find the probability that it does not reach safely is basically 11 - the probability that it reaches safely.
Let s denotes safe reaching and s’ denotes not reaching safe,
P(s)=13\Rightarrow P(s) = \dfrac{1}{3}
P(s)=113=23\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 =5C4(13)423{ = ^5}{C_4}{\left( {\dfrac{1}{3}} \right)^4}\dfrac{2}{3}
=5!4!×(54)!(13)423= \dfrac{{5!}}{{4! \times (5 - 4)!}}{\left( {\dfrac{1}{3}} \right)^4}\dfrac{2}{3}
=5!4!×1!×134×23= \dfrac{{5!}}{{4! \times 1!}} \times \dfrac{1}{{{3^4}}} \times \dfrac{2}{3}
=5×134×23= 5 \times \dfrac{1}{{{3^4}}} \times \dfrac{2}{3}
=5×235= 5 \times \dfrac{2}{{{3^5}}}
=1035= \dfrac{{10}}{{{3^5}}}
=1035= \dfrac{{10}}{{{3^5}}} --(1)
For 5 out of 5 case =5C5(13)5{ = ^5}{C_5}{\left( {\dfrac{1}{3}} \right)^5}
=5!5!×(55)!(13)5= \dfrac{{5!}}{{5! \times (5 - 5)!}}{\left( {\dfrac{1}{3}} \right)^5}
=5!5!×0!(13)5= \dfrac{{5!}}{{5! \times 0!}}{\left( {\dfrac{1}{3}} \right)^5}
=1×(13)5= 1 \times {\left( {\dfrac{1}{3}} \right)^5}
=(13)5= {\left( {\dfrac{1}{3}} \right)^5} --(2)
Now adding (1) and (2) we get the probability of at least four ships reach safely,
=1035+135=1135=11243= \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 11243\dfrac{{11}}{{243}} .
So, the correct answer is “ 11243\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.