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Probability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsProbability1 Mark
Question
A random variable has the following probability distribution:

X = xi 0 1 2 3 4 5 6 7
P(X = Xi) 0 2p 2p 3p p2 2p2 7p2 2p

  1. $\frac{1}{10}$

  2. $-1$

  3. $-\frac{1}{10}$

  4. $\frac{1}{5}$

✓

Answer

  1. $\frac{1}{10}$

Solution:

We know that the sum of probabilities in a probability distribution is always 1.

$\therefore$ P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) +P(X = 6) + P(X = 7) +P(X = 8) = 1

⇒ 0 + 2p + 2p + 3p + p2 + 2p2 + 7p2 + 2p = 1

⇒ 10p2 + 9p - 1 = 0

⇒ (10p - 1)(p + 1) = 0

$\Rightarrow\text{p}=\frac{1}{10}\text{ or }-1$ (Negleting -1 as the value of the probability cannot be negative)

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