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# A pair of dice is thrown 4 times. If getting a doublet is considered as success,(1) Find the probability of getting a doublet.(2) Hence, find the probability of two successes.

Last updated date: 17th Jul 2024
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Hint:- Use Bernoulli trials to find probability of getting two successes and probability of getting a favourable outcome is the ratio of number of favourable outcomes to the total number of outcomes.

(1)
For getting a doublet dice must be thrown 2 times.
As we know that dice has six numbers.
So, possible number of outcomes is,
S = { (1,1), (1,2), (1,3), ......... (6,4), (6,5), (6,6) }.

So, the total number of outcomes is n(S) = 6x6 = 36.
Let, outcomes with doublet be A.
So, A = { (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) }.
So, the number of outcomes with doublet is n(A) = 6.
As, we know that,
Probability of getting a favourable outcome = $\dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of outcomes}}}}$
So, probability of getting doublet P(A) $= {\text{ }}\dfrac{{{\text{n(A)}}}}{{{\text{n(S)}}}}$ $= {\text{ }}\dfrac{6}{{36}}{\text{ }} = {\text{ }}\dfrac{1}{6}$
Hence, the probability of getting a doublet will be $\dfrac{1}{6}$.

(2)
Let, B be the event of not getting successes.
Then, Probability of not getting success = 1 – Probability of getting success.
So, P(B) = 1 – P(A) = 1 – $\dfrac{1}{6}{\text{ = }}\dfrac{5}{6}$
Let, n be the number of times dice is thrown and that is 4.
Let X be the number of doublets.

So, throwing a pair of dice is a Bernoulli trial.
So, according to Bernoulli trial formula.
And, X has the binomial distribution.
P(X = x) = ${}^n{C_x}{q^{n - x}}{p^x}$

Where,
P(x) = probability of getting x successes.
x = number of successes.
n = number of times dice is thrown = 4
q = probability of not getting doublet = $\dfrac{5}{6}$
p = probability of getting doublet = $\dfrac{1}{6}$
And we need to find the probability of getting two successes.
So, x = 2

So, now putting the values in the binomial distribution. We will get,
P(x = 2) = ${}^4{C_2}{\left( {\dfrac{5}{6}} \right)^2}{\left( {\dfrac{1}{6}} \right)^2}$
P(x = 2) = $\dfrac{{4!}}{{2!*2!}}{\left( {\dfrac{5}{6}} \right)^2}{\left( {\dfrac{1}{6}} \right)^2}$
P(x = 2) = $6*\dfrac{{25}}{{36}}*\dfrac{1}{{36}} = \dfrac{{25}}{{216}}$
Hence, the probability of getting two successes will be $\dfrac{{25}}{{216}}$.

Note:- Whenever we come up with this type of problem then first, we will find the probability of getting favourable outcome and then we subtract that from 1, to get probability of not getting favourable outcome. After that we will apply Bernoulli trial to get the required probability.