MHT CET · Maths · Probability
Let X be a discrete random variable. The probability distribution of X is given below
\(\begin{array}{|l|c|c|c|} \hline \mathrm{X} & 30 & 10 & -10 \\ \hline \mathrm{P}(\mathrm{X}) & \frac{1}{5} & \mathrm{~A} & \mathrm{~B} \\ \hline \end{array}\)
and \(\mathrm{E}(\mathrm{X})=4\), then the value of AB is equal to
- A \(\frac{3}{10}\)
- B \(\frac{2}{15}\)
- C \(\frac{1}{15}\)
- D \(\frac{3}{20}\)
Answer & Solution
Correct Answer
(D) \(\frac{3}{20}\)
Step-by-step Solution
Detailed explanation
\( P(X=30) + P(X=10) + P(X=-10) = 1 \) \( \frac{1}{5} + A + B = 1 \Rightarrow A + B = \frac{4}{5} \) (1)
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