# The chance of India winning the toss is $\dfrac{3}{4}$. If it wins the toss, then its chance of victory is $\dfrac{4}{5}$ otherwise it is only $\dfrac{1}{2}$. Then the chance of India’s victory is:\begin{align} & \text{1}\text{.) }\dfrac{\text{1}}{\text{5}} \\ & \text{2}\text{.) }\dfrac{\text{3}}{\text{5}} \\ & \text{3}\text{.) }\dfrac{\text{3}}{\text{40}} \\ & \text{4}\text{.) }\dfrac{\text{29}}{\text{40}} \\ \end{align}

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Hint: Find out the probability of India’s victory based on two cases: that is India wins the toss and wins the match and the probability of India loses the toss and wins the match after that calculate the probability of India’s Victory to check which option is correct in the above given options.

We know that in different situations the measure of uncertainty is called probability. The ratio of favourable number of outcomes to the total number of outcomes is the classical theory of probability .In statistical concept the probability is based on observations and collection of facts but in modern reference in axiomatic approach of probability we use some universal truth concepts.
Basically there are three types of probabilities:
Theoretical Probability: It is based on the possible chances of something to happen.
Experimental Probability: It is based on the basis of the observations of an experiment.
Axiomatic Probability: In this probability a set of rules or axioms are set which applies to all types.
The formula of the probability of an event is:
$\text{probability}=\dfrac{\text{number of desired outcomes}}{\text{total number of favourable outcomes}}$
Or
$P(A)=\dfrac{n(A)}{n(S)}$
Where, $P(A)$ is the probability of an event $A$
$n(A)$ is the number of favourable outcomes
And $n(S)$ is the total number of possible outcomes of a set.
If the probability of occurring an event is $P(A)$ then the probability of not occurring an event is $P(A')=1-PA$
Now, according to the given question:
There are two conditions for the event to be happen:
Case1: India wins the toss and wins the match.
Case2: India loses the toss and wins the match.
Now we will find out the probability of the events:
Probability of India wins the toss:$\dfrac{3}{4}$
Probability of India wins the toss and got victory : $\dfrac{4}{5}$
If the probability of occurring an event is $P(A)$ then the probability of not occurring an event is $P(A')=1-PA$
Hence, Probability of India loses the toss:
$1-\dfrac{3}{4}$
$=\dfrac{4-3}{4}$
$=\dfrac{1}{4}$
Probability of India loses the toss and got victory:$\dfrac{1}{2}$
Now, $\text{P(victory)=P(India wins the toss and got victory)+P(India loses the toss and got victory)}$
$\text{P(victory)=(}\dfrac{3}{4}\times \dfrac{4}{5}\text{)+(}\dfrac{\text{1}}{4}\times \dfrac{\text{1}}{2}\text{)}$
$\text{=}\dfrac{3}{5}+\dfrac{1}{8}$
$=\dfrac{24+5}{40}$
$=\dfrac{29}{40}$

So, the correct answer is “Option 4”.

Note: In probability Null set $\phi$ and sample space $S$ also represent events because both are the subsets of $S$. Here $\phi$ represents an impossible event and $S$ represents a definite event. The subset $\phi$ of $S$ denotes an impossible event and the subset of $S$ of $S$itself denotes the sure event.