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Let X denote the number of colleges where you will apply after your results and P(X = x) denotes your probability of getting admission in x number of colleges. It is given that

\(P(X = x) = \begin{cases} kx\,\,,\text{if x = 0,1}\\ 2kx\,\,,\text{if x = 2}\\ k(5-x),\text{if x = 3 or 4}\\ 0,\text{if x >4} \end{cases} \)

Where k is a positive constant.

Find the value of k. Also, find the probability that you will get admission in

(i) exactly one college

(ii) at most 2 colleges

(iii) at least 2 colleges.

1 Answer

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Best answer

Given,

Our variable is X and from equation we see that it is taking values X = 0, 1, 2, 3, 4 …… (any whole number)

And it represents the number of colleges in which you are going to get admission.

According to equation given we have :

∴ P(X = 0) = k × 0 = 0

P(X = 1) = k ×1 = k

P(X = 2) = 2k × 2 = 4k

P(X = 3) = k(5 - 3) = 2k

P(X = 4) = k(5 - 4) = k

P(X > 4) = 0

As in question, it is not given either X is random variable or the given distribution is a probability distribution, we can’t apply any thing directly.

But we have a hint in the question,

As P(X = 0) = 0

It means the chance of not getting admission in any college 0

∴ admission is sure

Hence the sum of all probabilities must be equal to 1 as getting admission has become a sure event.

This makes the given distribution a probability distribution an X a random variable also.

∴ k + 4k + 2k + k = 1

8k = 1

k = 1/8

i) P(getting admission in exactly one college) = P(1) = k = 1/8

ii) P(getting admission in atmost 2 colleges) = P(X ≤ 2)

= P(X = 1)+P(X = 2)

= k + 4k = 5k = 5/8

iii) P(getting admission in atleast 2 colleges) = P(X ≥ 2)

= P(X = 2) + P(X = 3) + P(X = 4)

= 4k + 2k + k = 7k = 7/8

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