# The probability distribution function of a random variable X is given by where c > 0Find:i. c ii. P(X<2)iii. P(1<X≤2)

The key point to solve the problem:

If a probability distribution is given then as per its definition, Sum of probabilities associated with each value of a random variable of given distribution is equal to 1

i.e. ∑(pi) = 1

Given probability distribution is: For the probability distribution:

∑(pi) = 1   Note: It’s better to apply hit and trial method for solving cubic equation, just by putting 1 we get that it satisfies the equation. So ( c - 1 ) will be its one factor. After that divide the polynomial with c-1 which will give a quadratic factor which will be factorized easily.

But we can also proceed as the equation is solved if you can think upto that extent.   (c - 1){3c(c - 2)-1(c - 2)} = 0

(c - 1)(3c - 1)(c - 2) = 0

c = 1 or c = 2 or c = 1/3

As we are given that c > 0

All three values satisfy the given condition, so they must be the values.

BUT BE CAUTIOUS :

As we know that the probability of any event lies between 0 and 1, and we are using c to represent probabilities. So We need to check whether for these values probabilities are defined are valid or not.

Xi : 0 1 2

Pi : 3c3 4c-10c2 5c-1

As this distribution suggests c = 1 and c = 2 are going to make probabilities invalid.

So, c = 1/3 is the only solution for c … (i)

P(X<2) = 1 – P(X = 2)

= 1 – (5c – 1) = 2 – 5c

= 2 – 5/3 = 1/3

P(1 < X ≤ 3) = P(2)

= 5/3 – 1 = 2/3 ……(ii)

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