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

Hint: 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
\[n! = n(n - 1)(n - 2)(n - 4).......2.1\]