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COMEDK · Maths · 36. Probability

The probability distribution of a discrete random variable X is given as

X1242A3A5A
P(X)\(\dfrac{1}{2}\)\(\dfrac{1}{5}\)\(\dfrac{3}{25}\)K\(\dfrac{1}{25}\)\(\dfrac{1}{25}\)

\(\text { Then the value of } A \text { if } E(X)=2.94 \text { is }\)

  1. A \(\dfrac{1}{2}\)
  2. B \(\dfrac{1}{3}\)
  3. C 2
  4. D 3
Verified Solution

Answer & Solution

Correct Answer

(D) 3

Step-by-step Solution

Detailed explanation

The sum of probabilities in a probability distribution is equal to 1. \(\dfrac{1}{2} + \dfrac{1}{5} + \dfrac{3}{25} + K + \dfrac{1}{25} + \dfrac{1}{25} = 1\) \(\dfrac{25 + 10 + 6 + 2 + 2}{50} + K = 1\) \(\dfrac{45}{50} + K = 1 \Rightarrow K = 1 - \dfrac{9}{10} = \dfrac{1}{10}\)…