Question

The probability that a certain kind of component will survive a given shock is $\dfrac{3}{4}.$ The probability that among 5 components test at most 3 will survive is $\dfrac{47}{{{2}^{x}}},$ then what is the value of x.

Answer 1

Hint: Now, to find the cases of at most 3, we have to find the values of 0, 1, 2, 3 separately using the formula $P\left( x=r \right)={{\text{ }}^{n}}{{C}_{r}}{{\left( p \right)}^{r}}{{\left( q \right)}^{n-r}}$ where n is the total cases, r is the asked case, p is the probability and q = 1 – p and then add all of them.

Complete step-by-step answer:
In the question, we are said that a probability of certain kind of component will survive a given shock is $\dfrac{3}{4}.$ Now, we are given that the probability among 5 components tested at most 3 will survive in the given form of $\dfrac{47}{{{2}^{x}}}$ and we have to find the value of x.
We have to find the probability for at most three will survive which can be represented as $P\left( x\le 3 \right).$ We can write $P\left( x\le 3 \right)$ as the sum of the values of P(x = 0), P(x = 1), P(x = 2) and P(x = 3).
Now, we will find the values of P(x = 0), P(x = 1), P(x = 3) by using the formula, $P\left( x=r \right)={{\text{ }}^{n}}{{C}_{r}}{{\left( p \right)}^{r}}{{\left( q \right)}^{n-r}}.$
Here, n is the total number of components and r is the number of components chosen, p is the probability of each and q is 1 – p.
Now, we know that n is 5, p is $\dfrac{3}{4}$ and q is $\left( 1-\dfrac{3}{4} \right)$ or $\dfrac{1}{4}.$ So, we can substitute it. So, the value of P(x = 0) is $^{5}{{C}_{0}}{{\left( \dfrac{3}{4} \right)}^{0}}{{\left( \dfrac{1}{4} \right)}^{0}}$ which is equal to $\dfrac{1}{45}$ by using the fact that,
$^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$
The value of P(x = 1) is
${{\Rightarrow }^{5}}{{C}_{1}}{{\left( \dfrac{3}{4} \right)}^{1}}{{\left( \dfrac{1}{4} \right)}^{4}}$
$\Rightarrow \dfrac{^{5}{{C}_{1}}\times 3}{{{4}^{5}}}$
$\Rightarrow \dfrac{5\times 3}{{{4}^{5}}}$
$\Rightarrow \dfrac{15}{{{4}^{5}}}$
The value of P(x = 2) is
${{\Rightarrow }^{5}}{{C}_{2}}{{\left( \dfrac{3}{4} \right)}^{2}}{{\left( \dfrac{1}{4} \right)}^{3}}$
$\Rightarrow \dfrac{^{5}{{C}_{2}}\times {{3}^{2}}}{{{4}^{5}}}$
$\Rightarrow \dfrac{5!}{\left( 5-2 \right)!\times 2!}\times \dfrac{9}{{{4}^{5}}}$
$\Rightarrow \dfrac{10\times 9}{{{4}^{5}}}$
$\Rightarrow \dfrac{90}{{{4}^{5}}}$
And finally, the value of P(x = 3) is
${{\Rightarrow }^{5}}{{C}_{3}}{{\left( \dfrac{3}{4} \right)}^{3}}{{\left( \dfrac{1}{4} \right)}^{2}}$
$\Rightarrow \dfrac{^{5}{{C}_{3}}\times {{3}^{3}}}{{{4}^{5}}}$
$\Rightarrow \dfrac{5!}{\left( 5-3 \right)!\times 3!}\times \dfrac{27}{{{4}^{5}}}$
$\Rightarrow \dfrac{10\times 27}{{{4}^{5}}}$
$\Rightarrow \dfrac{270}{{{4}^{5}}}$
Now, we will find the values of $P\left( x\le 3 \right)$ by adding the values of P(x = 0), P(x = 1), P(x = 2) and P(x = 3) which are $\dfrac{1}{{{4}^{5}}},\dfrac{15}{{{4}^{5}}},\dfrac{90}{{{4}^{5}}},\dfrac{270}{{{4}^{5}}}.$ Hence, the value of $P\left( x\le 3 \right)$ is
$\dfrac{1}{{{4}^{5}}}+\dfrac{15}{{{4}^{5}}}+\dfrac{90}{{{4}^{5}}}+\dfrac{270}{{{4}^{5}}}$
$\Rightarrow \dfrac{1+15+90+270}{{{4}^{5}}}$
$\Rightarrow \dfrac{376}{{{4}^{5}}}$
$\Rightarrow \dfrac{8\times 47}{{{2}^{10}}}$
$\Rightarrow \dfrac{47}{{{2}^{7}}}$
So, the probability is $\dfrac{47}{{{2}^{7}}}.$ Now, on comparing the value, we get the value of x as 7.
Therefore, the value of x is 7.

Note: Generally, in many places, one can also see that in the formula, $P\left( x=r \right)={{\text{ }}^{n}}{{C}_{r}}{{\left( p \right)}^{r}}{{\left( 1-p \right)}^{n-r}}$ instead of q, (1 – p) is also there which is also the same as p + q = 1.