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Question: The probability that Sania wins Wimbledon tournaments finals is \[\dfrac{1}{3}\]. If Sania Mirza pla...

The probability that Sania wins Wimbledon tournaments finals is 13\dfrac{1}{3}. If Sania Mirza plays 33 round of Wimbledon final, the probability that she wins at least one round of match is:
A) 827\dfrac{8}{{27}}
B) 1927\dfrac{{19}}{{27}}
C) 1027\dfrac{{10}}{{27}}
D) 1727\dfrac{{17}}{{27}}

Explanation

Solution

The probability of an event calculates the chances of its occurrence and nonoccurrence. The probability always lies between 00 and 11. Here we will consider all the rounds as independent events and use Bernoulli trials. Bernoulli trials are independent trials in which we consider only two cases and these are success or failure.
Formula used:
The probability of xx successes in Bernoulli trials is given by the formula n!x!(n−x)!pxqn−x\dfrac{{n!}}{{x!\left( {n - x} \right)!}}{p^x}{q^{n - x}}.
Wherexx can take values from 0,1,2,.........n0,1,2,.........n pp denotes the successes, qq denotes failures and q=1−pq = 1 - p, nn denotes the number of total trials performed.

Complete step by step solution:
We have given that the probability of winning the tournament that is the value of pp is 13\dfrac{1}{3}.
So the probability that she will not win the tournament that is the value of qq is =1−13 = 1 - \dfrac{1}{3}
=23= \dfrac{2}{3}

The probability that she will win at least one round out of three rounds is the case in which she can win one round or two rounds or all the three rounds.
Now we will find the required probability using the formula n!x!(n−x)!pxqn−x\dfrac{{n!}}{{x!\left( {n - x} \right)!}}{p^x}{q^{n - x}} as follows,
P(x⩾1)=P(x=1)+P(x=2)+P(x=3)P\left( {x \geqslant 1} \right) = P\left( {x = 1} \right) + P\left( {x = 2} \right) + P\left( {x = 3} \right).
=3!1!(3−1)!(13)1(23)2+3!2!(3−2)!(13)2(23)1+3!3!(3−3)!(13)3(23)0= \dfrac{{3!}}{{1!\left( {3 - 1} \right)!}}{\left( {\dfrac{1}{3}} \right)^1}{\left( {\dfrac{2}{3}} \right)^2} + \dfrac{{3!}}{{2!\left( {3 - 2} \right)!}}{\left( {\dfrac{1}{3}} \right)^2}{\left( {\dfrac{2}{3}} \right)^1} + \dfrac{{3!}}{{3!\left( {3 - 3} \right)!}}{\left( {\dfrac{1}{3}} \right)^3}{\left( {\dfrac{2}{3}} \right)^0}
=49+29+127= \dfrac{4}{9} + \dfrac{2}{9} + \dfrac{1}{{27}}
=12+6+127= \dfrac{{12 + 6 + 1}}{{27}}
=1927= \dfrac{{19}}{{27}}

Therefore ,the probability that Sania will win at least one round is 1927\dfrac{{19}}{{27}}.
So, option (B) is the correct answer.

Note:
The probability of an event may be zero if the event is impossible to occur. Generally the probability of an event is equally likely to happen, that is it may be any number between 00 and 11. Bernoulli trials are very useful for finding the probability of events where there are more cases than usual.