Question

If the mean and the variance of a binomial variate X are 2 and 1 respectively, then the probability that X takes a value greater than or equal to one is
$\left( \text{a} \right)\text{ }\dfrac{1}{16}$
$\left( \text{b} \right)\text{ }\dfrac{9}{16}$
$\left( \text{c} \right)\text{ }\dfrac{3}{4}$
$\left( \text{d} \right)\text{ }\dfrac{15}{16}$

Answer 1

Hint: To solve the above question, we will first find out what is the binomial distribution of probability. Then we will use the concept that the mean of a binomial variate is np and the variance of a binomial variate is npq, where n is the total number of trials, p is the probability of success and q is the probability of failure. Using these formulas, we will find the value of n, p and q. Then we will find the value of P (x = 1), P (x = 2)……P (x = n) and all of them to get the required answer.

Complete step-by-step answer:
Before, solving the question given, we must know what binomial distribution of probability is. According to the binomial distribution of probability, the probability of getting exactly ‘r’ success in ‘n’ trials is given by the formula
$P\left( x=r \right)={{\text{ }}^{n}}{{C}_{r}}{{p}^{r}}{{q}^{n-r}}$
where p is the probability of getting success, r is the probability of failure and the sum of probability of success and probability of failure is equal to 1, i.e.
$p+q=1......\left( i \right)$
Now, the mean of any binomial variate is given by the formula,
$\text{Mean}=np$
In our case, it is given that mean of binomial variate x is 2. Then we will get,
$np=2.....\left( ii \right)$
Similarly, the variance of any binomial variate is given by the formula,
$\text{Variance}=npq$
In our case, it is given in the question that the variance of binomial variate x is 1. Thus, we will get,
$npq=1......\left( iii \right)$
On dividing (iii) by (ii), we will get,
$\Rightarrow \dfrac{npq}{np}=\dfrac{1}{2}$
$\Rightarrow q=\dfrac{1}{2}.....\left( iv \right)$
On putting the value of q from (iv) to (i), we will get,
$\Rightarrow p+\dfrac{1}{2}=1$
$\Rightarrow p=1-\dfrac{1}{2}$
$\Rightarrow p=\dfrac{1}{2}.....\left( v \right)$
On putting the value of p from (v) to (ii), we will get,
$\Rightarrow n\left( \dfrac{1}{2} \right)=2$
$\Rightarrow n=2\times 2$
$\Rightarrow n=4$
Now, as we have got the values of n, p and q, we will find out the probability that x takes a value greater than or equal to 1. Thus, we have to find P (x = 1), P (x = 2), P (x = 3) and P (x = 4) and then we will add all of them to get the required probability. Thus, we have,
$\text{Required Probability}=P\left( x=1 \right)+P\left( x=2 \right)+P\left( x=3 \right)+P\left( x=4 \right)$
$\Rightarrow \text{Required Probability}={{\text{ }}^{4}}{{C}_{1}}{{\left( \dfrac{1}{2} \right)}^{1}}{{\left( \dfrac{1}{2} \right)}^{4-1}}+{{\text{ }}^{4}}{{C}_{2}}{{\left( \dfrac{1}{2} \right)}^{2}}{{\left( \dfrac{1}{2} \right)}^{4-2}}+{{\text{ }}^{4}}{{C}_{3}}{{\left( \dfrac{1}{2} \right)}^{3}}{{\left( \dfrac{1}{2} \right)}^{4-3}}+{{\text{ }}^{4}}{{C}_{4}}{{\left( \dfrac{1}{2} \right)}^{4}}{{\left( \dfrac{1}{2} \right)}^{0}}$
$\Rightarrow \text{Required Probability}={{\text{ }}^{4}}{{C}_{1}}{{\left( \dfrac{1}{2} \right)}^{4}}+{{\text{ }}^{4}}{{C}_{2}}{{\left( \dfrac{1}{2} \right)}^{4}}+{{\text{ }}^{4}}{{C}_{3}}{{\left( \dfrac{1}{2} \right)}^{4}}+{{\text{ }}^{4}}{{C}_{4}}{{\left( \dfrac{1}{2} \right)}^{4}}$
$\Rightarrow \text{Required Probability}=\text{ }{{\left( \dfrac{1}{2} \right)}^{4}}\left[ ^{4}{{C}_{1}}+{{\text{ }}^{4}}{{C}_{2}}+{{\text{ }}^{4}}{{C}_{3}}+{{\text{ }}^{4}}{{C}_{4}} \right]$
$\Rightarrow \text{Required Probability}=\text{ }\dfrac{^{4}{{C}_{1}}+{{\text{ }}^{4}}{{C}_{2}}+{{\text{ }}^{4}}{{C}_{3}}+{{\text{ }}^{4}}{{C}_{4}}}{16}$
Now, the value of $^{n}{{C}_{r}}$ is given by $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}.$ Thus, we will get,
$\Rightarrow \text{Required Probability}=\text{ }\left( \dfrac{4!}{1!\times 3!} \right)+\left( \dfrac{4!}{2!\times 2!} \right)+\left( \dfrac{4!}{3!\times 1!} \right)+\left( \dfrac{4!}{0!\times 4!} \right)$
$\Rightarrow \text{Required Probability}=\text{ }\left( \dfrac{4\times 3\times 2\times 1}{3\times 2\times 1} \right)+\left( \dfrac{4\times 3\times 2\times 1}{2\times 1\times 2\times 1} \right)+\left( \dfrac{4\times 3\times 2\times 1}{3\times 2\times 1} \right)+\left( \dfrac{4\times 3\times 2\times 1}{1\times 4\times 3\times 2\times 1} \right)$
$\Rightarrow \text{Required Probability}=\dfrac{\left( \dfrac{4}{1} \right)+\left( \dfrac{4\times 3}{2} \right)+\left( \dfrac{4}{1} \right)+1}{16}$
$\Rightarrow \text{Required Probability}=\dfrac{4+6+4+1}{16}$
$\Rightarrow \text{Required Probability}=\dfrac{15}{16}$
Hence, option (d) is the right answer.

Note: We can also approach the question in the following way after finding the values of n, p and q. We have to find the probability that x takes a value greater than or equal to 1. For this, we will first find P (x = 0) and subtract it from 1. Thus, we will get,
$P\left( x\ge 1 \right)=1-P\left( x=0 \right)$
$\Rightarrow P\left( x\ge 1 \right)=1-{{\text{ }}^{4}}{{C}_{0}}{{\left( \dfrac{1}{2} \right)}^{0}}{{\left( \dfrac{1}{2} \right)}^{4-0}}$
$\Rightarrow P\left( x\ge 1 \right)=1-1\times \dfrac{1}{16}$
$\Rightarrow P\left( x\ge 1 \right)=1-\dfrac{1}{16}$
$\Rightarrow P\left( x\ge 1 \right)=\dfrac{15}{16}$