question_answer

Two dice are thrown simultaneously. Let X denotes the number of sixes. Find the probability distribution of X.

OR

A and B are two candidates seeking admission in a college. The probability that A is selected, is 0.7 and the probability that exactly one of them is selected, is 0.6. Find the probability that B is selected.

Answer:

Clearly, P (getting six on a die) \[=\frac{1}{6}\]

And P (not getting six on a die) \[=1-\frac{1}{6}=\frac{5}{6}\]

Since, two dice are thrown and X denotes the number of sixes, so X can take values 0 (no six), 1 (one six) and 2 (two sixes).

Probability of getting no six,

\[P\,(X=0)=\frac{5}{6}\times \frac{5}{6}=\frac{25}{36}\]

Probability of getting one six,

\[P\,(X=1)=2\times \frac{5}{6}\times \frac{1}{6}=\frac{10}{36}\]

Probability of getting two sixes,

\[P\,(X=2)=\frac{1}{6}\times \frac{1}{6}=\frac{1}{36}\]

Thus, the required probability distribution is

X 0 1 2 P(X) \[\frac{25}{36}\] \[\frac{10}{36}\] \[\frac{1}{36}\]

OR

Let \[{{E}_{1}}\] be the event that candidate A is selected and \[{{E}_{2}}\] be the event that candidate B is selected.

Then, we have

\[P({{E}_{1}})=0.7\] and P(exactly one of A, B is selected) = 0.6

or P(A is selected, B is not selected or B is selected, A is not selected) = 0.6

or         \[P(({{E}_{1}}\cap {{\bar{E}}_{2}})\cup ({{\bar{E}}_{1}}\cap {{E}_{2}}))=0.6\]

Now, consider

\[P({{E}_{1}}\cap {{\bar{E}}_{2}})\cup ({{\bar{E}}_{1}}\cap {{E}_{2}})=0.6\]

\[\therefore \]      \[P({{E}_{1}}\cap {{\bar{E}}_{2}})+P({{\bar{E}}_{1}}\cap {{E}_{2}})=0.6\]

[\[\because \] events \[{{E}_{1}}\cap {{\bar{E}}_{2}}\] and \[{{\bar{E}}_{1}}\cap {{E}_{2}}\]

are mutually exclusive]

\[\therefore \]     \[P({{E}_{1}}).P({{\bar{E}}_{2}})+P({{\bar{E}}_{1}}).P({{E}_{2}})=0.6\]

[\[\because \] \[{{E}_{1}}\] and \[{{E}_{2}}\] are independent events]

\[\Rightarrow \]   \[(0.7)P({{\bar{E}}_{2}})+(0.3)P({{E}_{2}})=0.6\]

\[[\because \,\,\,P({{\bar{E}}_{1}})=1-P({{E}_{1}})=1-0.7=0.3]\]

\[\Rightarrow \]   \[0.7\,(1-P({{E}_{2}}))+0.3P({{E}_{2}})=0.6\]

\[\Rightarrow \]   \[0.7-0.7P({{E}_{2}})+0.3P({{E}_{2}})=0.6\]

\[\Rightarrow \]   \[0.4P({{E}_{2}})=0.1\] \[\Rightarrow \] \[P({{E}_{2}})=\frac{1}{4}\]