Probability — Maths STD 12 Science — Question

1. Since, the sum of all the probabilities of a distribution is 1.

∴ P(X = 0) + P(X = 1) + …. + P(X = 7) = 1

⇒ 0 + k + 2k + 2k + 3k + k² + 2k² + 7k² + k = 1

⇒ 10k² + 9k - 1 = 0

⇒ (10k - 1) (k + 1) = 0

⇒ 10k - 1 = 0 or k + 1 = 0

 $\Rightarrow\ \text{k}=\frac{1}{10}$ or k = - 1

Since, k ≥ 0, therefore k = − 1 is not possible.

$\therefore\ \text{k}=\frac{1}{10}$

2. P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

= 0 + k + 2k

$=3\text{k}=3\times\frac{1}{10}=\frac{3}{10}$

3. P(X > 6) = P(X = 7)

$=7\text{k}^2+\text{k}=7\Big(\frac{1}{10}\Big)^2+\frac{1}{10}=\frac{17}{100}$

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

= k + 2k = 3k = $\frac{3}{10}$