Book: RD Sharma - Mathematics (Volume 2)
Chapter: 33. Binomial Distribution
Subject: Maths - Class 12th
Q. No. 50 of Exercise 33.1
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Find the probability that in 10 throws of a fair die a score which is a multiple of 3 will be obtained in at least 8 of the throws.
Given that, obtaining multiple of 3 in a throw of a die is a success.
There are total 10 throws of a fair die.
Let p be the probability of getting multiple of 3 in a throw of a die.
Since there can be a total 6 outcomes in a throw of a die. That is, {1, 2, 3, 4, 5, 6}.
And a multiple of 3 in a die is {3}, {6}.
Then, let q be the probability of not getting a multiple of 3 in a throw of a die.
And we know, p + q = 1
⇒ q = 1 – p
Let X be a random variable representing a number of successes (getting multiple of 3 in die) out of 10 throws of a die.
Then, the probability of getting r successes out of n throws of die is given by,
P (X = r) = nCrprqn-r
Here n = 10.
Now, put values of n, p, and q in the above equation.
…(i)
We need to find the probability of getting a multiple of 3 in at least 8 of the throws out of 10 throws of a fair die.
It is given,
Probability = P (X ≥ 8)
This can be written as,
P (X ≥ 8) = P (X = 8) + P (X = 9) + P (X = 10)
Just out r = 8, 9, 10 in equation (i) to find the value of P (X = 8), P (X = 9), P (X = 10) respectively, then substitute in the above equation.
Thus, the probability of getting a multiple of 3 in at least 8 of the throws out of 10 throws of a fair die is 201/310.
