Question

An irregular 6 faced dice is such that the probability that it gives 3 even numbers in 5 throws is twice the probability that it gives 2 even numbers 5 throws. How many sets of exactly 5 trials can be expected to give no even number out of 2500 sets?

Answer 1

Answer:

10

Step-by-step explanation:

Hi,

Let X be the random variable which denotes the number of even number in 5 throws

Let the probability of getting even = p

So, the probability of getting odd will be 1 - p

Given that P( X = 3) = 2 P(X = 2)

But P(X = r) is given by nCr(p)^r(q)^(n-r)

Hence,

5C3p³(1-p)² = 2* 5C2p²(1-p)³

⇒ p = 2(1-p)

⇒3p = 2

⇒p = 2/3

Thus, probability of getting an even number is 2/3.

In a set of exactly 5 trials, the probability of getting '0' even numbers will be

5C0(p)⁰(1-p)⁵ = 1/243. So,

So, out of 2500 sets, the total expected  number of sets of 5 having no even number will be 2500*1/243 ≈ 10.28

Hence not more than 10 set are expected to give no even numbers.

Hope, it helped !