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MHT CET · Maths · Probability

A fair die is tossed twice in succession. If \(\mathrm{X}\) denotes the number of fours in two tosses, then the probability distribution of \(\mathrm{X}\) is given by

  1. A \(\begin{array}{|c|c|c|c|}
    \hline \mathrm{X}=x_{\mathrm{i}} & 0 & 1 & 2\mathrm{P}_{\mathrm{i}} & \frac{1}{36} & \frac{25}{36} & \frac{5}{18}\\ \hline\end{array}\)
  2. B \(\begin{array}{|c|c|c|c|}
    \hline \mathrm{X}=x_{\mathrm{i}} & 0 & 1 & 2\mathrm{P}_{\mathrm{i}} & \frac{25}{36} & \frac{1}{36} & \frac{5}{18} \\ \hline\end{array}\)
  3. C \(\begin{array}{|c|c|c|c|}
    \hline \mathrm{X}=x_{\mathrm{i}} & 0 & 1 & 2\mathrm{P}_{\mathrm{i}} & \frac{25}{36} & \frac{5}{18} & \frac{1}{36} \\ \hline \end{array}\)
  4. D \(\begin{array}{|c|c|c|c|}
    \hline \mathrm{X}=x_{\mathrm{i}} & 0 & 1 & 2\mathrm{P}_{\mathrm{i}} & \frac{5}{18} & \frac{1}{36} & \frac{25}{36} \\ \hline \end{array}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\begin{array}{|c|c|c|c|}
\hline \mathrm{X}=x_{\mathrm{i}} & 0 & 1 & 2\mathrm{P}_{\mathrm{i}} & \frac{25}{36} & \frac{5}{18} & \frac{1}{36} \\ \hline \end{array}\)

Step-by-step Solution

Detailed explanation

A fair die is tossed twice in succession.
\(\therefore \) Sample space (S)
\(\begin{aligned} = & (1,1),(1,2),(1,3),(1,4),(1,5),(1,6), \\ & (2,1),(2,2),(2,3),(2,4),(2,5),(2,6), \\ & (3,1),(3,2),(3,3),(3,4),(3,5),(3,6), \\ & (4,1),(4,2),(4,3),(4,4),(4,5),(4,6), \\ & (5,1),(5,2),(5,3),(5,4),(5,5),(5,6), \\ & (6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\} \end{aligned}\)
\(X:\) Number of fours in two tosses.
\(\therefore \) Possible values of \(\mathrm{X}\) are: \(0,1,2\).
\(\therefore \) Probability distribution of \(\mathrm{X}\) is as follows:
\(\begin{array}{|c|c|c|c|} \hline \mathrm{X}=x_{\mathrm{i}} & 0 & 1 & 2 \\ \hline \mathrm{P}_{\mathrm{i}} & \frac{25}{36} & \frac{10}{36}=\frac{5}{18} & \frac{1}{36} \\ \hline \end{array}\)