Question

India and Pakistan play a 5 test match series of cricket, the probability that India
Wins at least three matches is
$\begin{align}
  & \left( a \right)\dfrac{1}{2} \\
 & \left( b \right)\dfrac{3}{5} \\
 & \left( c \right){{\left( \dfrac{1}{2} \right)}^{3}} \\
 & \left( d \right)\text{None of these} \\
\end{align}$

Answer 1

Hint: First calculate all probabilities with India winning greater than or equal to 3. Now add these probabilities to get the total probability. Look at definitions of rule of sum, Rule of product applies to the concept which is suitable here. Total probability is the required result.

Complete step-by-step answer:
Given conditions for which we need to calculate the probability. Out of 5 matches India must win 3. As probability we take P of India =$\dfrac{1}{2}$ , P of Pakistan =$\dfrac{1}{2}$
Case (1): India winning 3 matches out of 5 matches held. 5 matches can be said as:
I I I P P
I is India and P represents Pakistan. As equal as probability we get probability as:
$P=\dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}\text{=}\dfrac{1}{{{2}^{5}}}$
For each case the probability is this. Total cases are total arrangement of 5 letters with 3 of one kind 2 of other arrangements of n things with a of one kind b of other kind is
$\dfrac{a!}{a!b!}.........\left( 1 \right)$
By applying this we get the probability as
=$\dfrac{1}{{{2}^{5}}}\times \dfrac{5!}{3!2!}......\left( 2 \right)$
Case (2): India winning 4 matches out of 5 matches’ held. 5 matches can be written in form of
 I I I I P
It is India and P is Pakistan. As equal probability each has $\dfrac{1}{2}$. We get probability as given here:
$P =\dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}\times \dfrac{1}{2}=\dfrac{1}{{{2}^{5}}}$.
For each case this is probability. Total cases are arrangements of 5 letters of which 4 are of the same kind.
Total arrangements =$\dfrac{5!}{4!}$
Probability=$\dfrac{1}{{{2}^{5}}}\times \dfrac{5!}{4!}.........\left( 3 \right)$
Case (3): India winning 5 matches out of 5 matches held. 5 matches are written in the form of
 I I I I I
 Only I arranged possible equal probability implies each by $\dfrac{1}{2}$ so, total probability
=$\dfrac{1}{{{2}^{5}}}\times 1........\left( 4 \right)$
Rule of sum: In combinatorics, the rule of addition principle is basic counting principle. It is simply defined as, if there are A ways of doing P work and B ways of doing Q work. Given P, Q words cannot be done together. Total number of ways to do both P, Q are given by (A+B) ways.
Rule of Product: In combinatorics, the rule of product or multiplication principle is basic counting principle. It is simply defined as, if there are A ways of doing P work and B ways of doing Q work. Given P, Q works can be done at a time. Total number of ways to do both P, Q works are given by (A.B) ways.
We know all cases are separate; they cannot occur at the same time. So, by the definitions above, we need a sum rule. By applying sum rule, we get the probability P as:
P= equation (2) + equation (3) + equation (4)
By substituting values, we get the value of P as:
$P=\dfrac{1}{{{2}^{5}}}\times \dfrac{5!}{3!2!}+\dfrac{1}{{{2}^{5}}}\times \dfrac{5!}{4!}\text{+}\dfrac{1}{{{2}^{5}}}\times 1$
By applying factorial definition, we can write equation as:
\[P=\dfrac{1}{{{2}^{5}}}\times \dfrac{5\times 4}{2}+\dfrac{1}{{{2}^{5}}}\times 5+\dfrac{1}{{{2}^{5}}}\]
By simplifying and taking $\dfrac{1}{{{2}^{5}}}$ common, we can write equation as
$P=\dfrac{1}{{{2}^{5}}}\left( \text{10+5+1} \right)=\dfrac{1}{{{2}^{5}}}16=\dfrac{16}{32}$
By simplifying we get final answers as P=$\dfrac{1}{2}$
Therefore option (a) is correct.

Note: Be careful while calculating the number of cases. If you miss any number then the whole answer might go wrong. When you apply the sum rule add then carefully. Generally students confuse and apply product rules. So, read the definitions carefully and apply them properly.