MHT CET · Maths · Probability
A fair die is tossed twice in succession. If \(\mathrm{X}\) denotes the number of sixes in two tosses, then the probability distribution of \(\mathrm{X}\) is given by
- A \(\begin{array}{|l|c|c|c|}
\hline \mathrm{X}=x & 0 & 1 & 2 \mathrm{P}(\mathrm{X}=x) & \frac{25}{36} & \frac{1}{36} & \frac{5}{18} \\ \hline \end{array}\) - B \(\begin{array}{|l|c|c|c|}
\hline \mathrm{X}=x & 0 & 1 & 2 \mathrm{P}(\mathrm{X}=x) & \frac{5}{18} & \frac{1}{36} & \frac{25}{36} \\ \hline\end{array}\) - C \(\begin{array}{|l|c|c|c|}
\hline \mathrm{X}=x & 0 & 1 & 2\mathrm{P}(\mathrm{X}=x) & \frac{25}{36} & \frac{5}{18} & \frac{1}{36} \\ \hline\end{array}\) - D \(\begin{array}{|l|c|c|c|}
\hline \mathrm{X}=x & 0 & 1 & 2\mathrm{P}(\mathrm{X}=x) & \frac{5}{18} & \frac{25}{36} & \frac{1}{36} \\ \hline\end{array}\)
Answer & Solution
Correct Answer
(C) \(\begin{array}{|l|c|c|c|}
\hline \mathrm{X}=x & 0 & 1 & 2\mathrm{P}(\mathrm{X}=x) & \frac{25}{36} & \frac{5}{18} & \frac{1}{36} \\ \hline\end{array}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{X}\) can take values 0,1 and 2 .
\(P(X=0)=\) Probability of not getting six \(=\frac{25}{36}\)
\(\mathrm{P}(\mathrm{X}=1)=\) Probability of getting one six
\(=\frac{10}{36}=\frac{5}{18}\)
\(P(X=2)=\) Probability of getting two sixes \(=\frac{1}{36}\)
The probability distribution of \(\mathrm{X}\) is
\(\begin{array}{|l|c|c|c|} \hline \mathrm{X}=x & 0 & 1 & 2 \\ \hline \mathrm{P}(\mathrm{X}=x) & \frac{25}{36} & \frac{5}{18} & \frac{1}{36} \\ \hline \end{array}\)
\(P(X=0)=\) Probability of not getting six \(=\frac{25}{36}\)
\(\mathrm{P}(\mathrm{X}=1)=\) Probability of getting one six
\(=\frac{10}{36}=\frac{5}{18}\)
\(P(X=2)=\) Probability of getting two sixes \(=\frac{1}{36}\)
The probability distribution of \(\mathrm{X}\) is
\(\begin{array}{|l|c|c|c|} \hline \mathrm{X}=x & 0 & 1 & 2 \\ \hline \mathrm{P}(\mathrm{X}=x) & \frac{25}{36} & \frac{5}{18} & \frac{1}{36} \\ \hline \end{array}\)
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