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: 23rd Mar 2023
•
Total views: 308.7k
•
Views today: 3.86k
Answer
308.7k+ views
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.
(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.
Recently Updated Pages
If ab and c are unit vectors then left ab2 right+bc2+ca2 class 12 maths JEE_Main

A rod AB of length 4 units moves horizontally when class 11 maths JEE_Main

Evaluate the value of intlimits0pi cos 3xdx A 0 B 1 class 12 maths JEE_Main

Which of the following is correct 1 nleft S cup T right class 10 maths JEE_Main

What is the area of the triangle with vertices Aleft class 11 maths JEE_Main

KCN reacts readily to give a cyanide with A Ethyl alcohol class 12 chemistry JEE_Main

Trending doubts
What was the capital of Kanishka A Mathura B Purushapura class 7 social studies CBSE

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Tropic of Cancer passes through how many states? Name them.

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

Name the Largest and the Smallest Cell in the Human Body ?
