
In an over of $6$ balls bowled by a bowler, the probability that he will get exactly three wickets in consecutive balls is (assume that the probability of getting a wicket is $0.5$).
A)$\dfrac{4}{{{4^4}}}$
B)$\dfrac{4}{{{4^5}}}$
C)$\dfrac{1}{{{2^4}}}$
D)$\dfrac{1}{{32}}$
Answer
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Hint: Write down all the possible outcomes in which the bowler can get exactly three wickets in consecutive balls. Since the probability of getting a wicket is $0.5$, the probability of not getting the wicket will be $\left( {1 - 0.5 = 0.5} \right)$.
Complete step-by-step answer:
Let us write down the cases for getting three consecutive wickets. Let getting a wicket be denoted by W and not getting a wicket be denoted by N.
So, the cases are
WWWNNN, NWWWNN, NNWWWN, NNNWWW.
So, we can clearly see that the favourable number of outcomes is $4$.
Since there are $6$ balls in one over, and it can have only two possible outcomes, that is, either it hits a wicket or it does not hit the wicket, hence,
The total number of outcomes can be $2 \times 2 \times 2 \times 2 \times 2 \times 2$
(because there are six balls which can have two outcomes.)
Total outcomes $ = {2^6}$
Favourable outcomes $ = 4 = {2^2}$
Probability = favourable outcomes/total outcomes
$ = \dfrac{{{2^2}}}{{{2^6}}} = \dfrac{1}{{{2^4}}}$
Hence, C is the correct answer.
Note: In these types of questions, the first and foremost task is to count the number of ways in which the given event may happen. Then calculate the total number of outcomes and then you could easily calculate the probability.
Complete step-by-step answer:
Let us write down the cases for getting three consecutive wickets. Let getting a wicket be denoted by W and not getting a wicket be denoted by N.
So, the cases are
WWWNNN, NWWWNN, NNWWWN, NNNWWW.
So, we can clearly see that the favourable number of outcomes is $4$.
Since there are $6$ balls in one over, and it can have only two possible outcomes, that is, either it hits a wicket or it does not hit the wicket, hence,
The total number of outcomes can be $2 \times 2 \times 2 \times 2 \times 2 \times 2$
(because there are six balls which can have two outcomes.)
Total outcomes $ = {2^6}$
Favourable outcomes $ = 4 = {2^2}$
Probability = favourable outcomes/total outcomes
$ = \dfrac{{{2^2}}}{{{2^6}}} = \dfrac{1}{{{2^4}}}$
Hence, C is the correct answer.
Note: In these types of questions, the first and foremost task is to count the number of ways in which the given event may happen. Then calculate the total number of outcomes and then you could easily calculate the probability.
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