Question

The probability of the safe arrival of one ship out of 5 is $\dfrac{1}{5}$. What is the probability of the safe arrival of at least 3 ships?
A. $\dfrac{1}{{31}}$
B. $\dfrac{3}{{52}}$
C. $\dfrac{{181}}{{3125}}$
D. $\dfrac{{184}}{{3125}}$

Answer 1

Hint: We use the Bernoulli’s trial, which is given by; ${\text{P}}\left( {{\text{Success}} = x} \right) = {}^n{{\text{C}}_x}{p^x}{q^{n - x}}$ , where the value of x varies from 0 to n. Also, it is important to note that the relation between p and q is that their sum is equal to 1.

Complete step-by-step solution
Let us read the question and consider the given things. It is given that the probability of the safe arrival is $\dfrac{1}{5}$. We need to find the probability of the safe arrival of at least 3 ships.
We use the method or the formula of Bernoulli’s trial.

First let that the probability of the safe arrival which is, $\dfrac{1}{5}$, be equal to p. Thus, we find the value of q by using the relation, $p + q = 1$ .
$\Rightarrow \dfrac{1}{5} + q = 1 \\\ \Rightarrow q = 1 - \dfrac{1}{5} \\\ \Rightarrow q = \dfrac{4}{5} \\\$

Also, we see that it is given in the question that the value of n is 5. Thus, the probability of the safe arrival of at least 3 ships is given by:
${\text{P}}\left( 3 \right) + {\text{P}}\left( 4 \right) + {\text{P}}\left( 5 \right)$

We use all the values in the Bernoulli’s trial as follows;

$$\Rightarrow {}^5{{\text{C}}_3}{\left( {\dfrac{1}{5}} \right)^3}{\left( {\dfrac{4}{5}} \right)^{5 - 3}} + {}^5{{\text{C}}_4}{\left( {\dfrac{1}{5}} \right)^4}{\left( {\dfrac{4}{5}} \right)^{5 - 4}} + {}^5{{\text{C}}_5}{\left( {\dfrac{1}{5}} \right)^5}{\left( {\dfrac{4}{5}} \right)^{5 - 5}} \\\ = {}^5{{\text{C}}_3}{\left( {\dfrac{1}{5}} \right)^3}{\left( {\dfrac{4}{5}} \right)^2} + {}^5{{\text{C}}_4}{\left( {\dfrac{1}{5}} \right)^4}{\left( {\dfrac{4}{5}} \right)^1} + {}^5{{\text{C}}_5}{\left( {\dfrac{1}{5}} \right)^5}{\left( {\dfrac{4}{5}} \right)^0} \\\ = 10 \times \dfrac{1}{{125}} \times \dfrac{{16}}{{25}} + 5 \times \dfrac{1}{{625}} \times \dfrac{4}{5} + 1 \times \dfrac{1}{{3125}} \times 1 \\\ = \dfrac{{160}}{{3125}} + \dfrac{{20}}{{3125}} + \dfrac{1}{{3125}} \\\ = \dfrac{{181}}{{3125}} \\\$$

Thus, option (C) is the correct option.

Note: Carefully observe what you need to find and what is given to you. Interchanging the two can lead to a wrong answer. Get familiar with the concept of Bernoulli’s trial in order to make the right choice for the success and hence apply the correct Bernoulli’s trial to that corresponding success. Avoid making calculation mistakes.