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A random variable X has the following probability distribution

X012345678
P(X)a3a5a7a9a11a13a15a17a

Then the values of 'a' and P(0 < X < 5) respectively are

Solution

✅ Correct Option: 1

In any probability distribution, the sum of all probabilities must equal 1.

P(X=0)+P(X=1)+P(X=2)+...+P(X=8)=1P(X=0) + P(X=1) + P(X=2) + ... + P(X=8) = 1

Substituting the values from the table:

a+3a+5a+7a+9a+11a+13a+15a+17a=1a + 3a + 5a + 7a + 9a + 11a + 13a + 15a + 17a = 1

81a=181a = 1

a=181a = \dfrac{1}{81}


The notation 0<X<50 < X < 5 means XX is greater than 0 and less than 5.

XX can take values: 1,2,3,41, 2, 3, 4

P(0<X<5)=P(X=1)+P(X=2)+P(X=3)+P(X=4)P(0 < X < 5) = P(X=1) + P(X=2) + P(X=3) + P(X=4)

P(0<X<5)=3a+5a+7a+9aP(0 < X < 5) = 3a + 5a + 7a + 9a

P(0<X<5)=24aP(0 < X < 5) = 24a

Substituting a=181a = \dfrac{1}{81}:

P(0<X<5)=24×181P(0 < X < 5) = 24 \times \dfrac{1}{81}

P(0<X<5)=2481P(0 < X < 5) = \dfrac{24}{81}

P(0<X<5)=827P(0 < X < 5) = \dfrac{8}{27}


Therefore:

a=181a = \dfrac{1}{81} and P(0<X<5)=827P(0 < X < 5) = \dfrac{8}{27}

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