Question

The probability that Dhoni will hit a century in every ODI match he plays is $\dfrac{1}{5}$. If he plays 6 matches in World Cup 2011, the probability that he will score 2 centuries is:
A.$\dfrac{{768}}{{3125}}$
B.$\dfrac{{2357}}{{3125}}$
C.$\dfrac{{2178}}{{3125}}$
D.$\dfrac{{412}}{{3125}}$

Answer 1

Here, we have to find the probability that Dhoni will score 2 centuries. First we will find the probability that he will not hit a century using the property of the sum of an event and the complimentary event. Then we will substitute all the probabilities in the formula of Binomial distribution to find the probability that he will score 2 centuries. Binomial Distribution is the probability distribution of the number of successes in a sequence of independent trials.

Formula Used:
We will use the following formula:
1.Sum of an event and a complementary event $p + q = 1$
2.Binomial Probability Distribution is given by the formula ${P_r} = {}^n{C_x}{p^x}{q^{n - x}}$, where $P$ is the binomial Probability, $x$ is the number of times that the outcome occurs in $n$-trials, $n$ is the number of trials, ${}^n{C_x}$ is the number of combinations, $p$ is the probability of success, $q$ is the probability of failures.
3.Combination is given by the formula ${}^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$.

Complete step-by-step answer:
We are given that the probability that Dhoni will hit a century in every ODI match played is $\dfrac{1}{5}$.
So, the Probability of success, $p = \dfrac{1}{5}$
We can find the probability that he will not hit a century by sum of an event and a complementary event . So,
$p + q = 1$
Substituting $p = \dfrac{1}{5}$ in the above equation, we get
$\Rightarrow \dfrac{1}{5} + q = 1$
Subtracting $\dfrac{1}{5}$ from both the sides, we get
$\Rightarrow q = 1 - \dfrac{1}{5}$
On cross-multiplication, we get
$\Rightarrow q = \dfrac{{5 - 1}}{5}$
$\Rightarrow q = \dfrac{4}{5}$
So, the probability that he will not score a century is $\dfrac{4}{5}$.
Total matches played by Dhoni $= 6$
Since we are given the probability of success or failure, we use the binomial probability distribution.
Substituting $n = 6,r = 2,p = \dfrac{1}{5}$ and $q = \dfrac{4}{5}$ in the formula ${P_r} = {}^n{C_x}{p^x}{q^{n - x}}$, we get
$\Rightarrow {P_2} = {}^6{C_2} \times {\left( {\dfrac{1}{5}} \right)^2} \times {\left( {\dfrac{4}{5}} \right)^{6 - 2}}$
$\Rightarrow {P_2} = {}^6{C_2} \times {\left( {\dfrac{1}{5}} \right)^2} \times {\left( {\dfrac{4}{5}} \right)^4}$
Now using the formula of Combination ${}^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$, we get
$\Rightarrow {P_2} = \dfrac{{6!}}{{(6 - 2)!2!}} \times {\left( {\dfrac{1}{5}} \right)^2} \times {\left( {\dfrac{4}{5}} \right)^4}$
Subtracting the terms in the bracket, we get
$\Rightarrow {P_2} = \dfrac{{6!}}{{4!2!}} \times {\left( {\dfrac{1}{5}} \right)^2} \times {\left( {\dfrac{4}{5}} \right)^4}$
Simplifying the factorial, we get
$\Rightarrow {P_2} = 5 \times 3 \times {\left( {\dfrac{1}{5}} \right)^2} \times {\left( {\dfrac{4}{5}} \right)^4}$
Thus by simplification, we get
$\Rightarrow {P_2} = 3 \times \left( {\dfrac{1}{5}} \right) \times {\left( {\dfrac{4}{5}} \right)^4}$
$\Rightarrow {P_2} = 3 \times \dfrac{{256}}{{3125}}$
Multiplying the terms, we get
$\Rightarrow {P_2} = \dfrac{{768}}{{3125}}$
Therefore, the probability that he will score 2 centuries is $\dfrac{{768}}{{3125}}$.
Hence, option A is the correct answer.

Note: Here, we are using the binomial distribution since we have only two outcomes either success or failure. Binomial distribution should satisfy the conditions that include the number of trials are constant. Each observation is independent from all the other trials since it has no effect on the other. The probability of success is the same for all the trials. We should know that the binomial distribution should be used only if it satisfies all the conditions. Thus Binomial distribution is used to find only the probability of success in the independent trials.