Question

India plays two matches each with West Indies and Australia. In any match the probabilities of India getting points $0, 1, 2$ are $0.45, 0.05, 0.50$ respectively. Assuming that the outcomes are independent, the probability of India getting at least $7$ point is
(A) $0.8750$
(B) $0.0875$
(C) $0.0625$
(D) $0.0250$

Answer 1

In this question, we have the probabilities of getting specific points. First we need to find out the number of possible cases for getting at least $7$ point. Then we will calculate the possible outcomes for each case. Then we can easily find out the required solution.

Complete step-by-step answer:
It is given that India plays two matches each with West Indies and Australia.
In any match the probabilities of India getting points $0, 1, 2$ are $0.45, 0.05, 0.50$ respectively.
Also given that, we assume the outcomes are independent.
We need to find out the probability of India getting at least $7$ point.
Since there are only four matches played by India.
Thus, India can get maximum $8$points.
We need to find out the points $\geqslant 7$
For the case where India will get $7$ point:
India will get $2$ in $3$ matches and $1$ in one match.
Then the probability of getting $2$ in each of the $3$ matches and $1$ in one match =$^4{C_3}{\left( {0.5} \right)^3}\left( {0.05} \right) = \dfrac{{4!}}{{3!1!}} \times 0.125 \times 0.05$
$\Rightarrow 4 \times 0.00625$
On multiply we get,
$\Rightarrow 0.025$
Also, for the case where India will get $8$ point:
India will get $2$ in each of four matches.
Then the probability of getting $2$ in each of the four matches,
$^4{C_4}{\left( {0.5} \right)^4} = \dfrac{{4!}}{{4!0!}} \times 0.0625$
$\Rightarrow 1 \times 0.0625$
On multiply we get,
$= 0.0625$
Now, we can find the probability of India getting at least $7$ point
$\Rightarrow$ Probability of getting exactly $7$ point + Probability of getting exactly $8$ point
$\Rightarrow 0.025 + 0.0625$
Let us adding the term and we get,
$= 0.0875$

Therefore (B) is the correct option.

Note: In mathematics, a combination is a selection of items from a collection, such that the order of selection does not matter.
For a combination,
$C\left( {n,{\text{ }}r} \right){ = ^n}{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$
Here, factorial n is denoted by $n!$ and we can define by
$n! = n(n - 1)(n - 2)(n - 4).......2.1$