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] \[\Rightarrow \] \[(0.7)P({{\bar{E}}_{2}})+(0.3)P({{E}_{2}})=0.6\] \[[\because \,\,\,P({{\bar{E}}_{1}})=1-P({{E}_{1}})=1-0.7=0.3]\] \[\Rightarrow \] \[0.7\,(1-P({{E}_{2}}))+0.3P({{E}_{2}})=0.6\] \[\Rightarrow \] \[0.7-0.7P({{E}_{2}})+0.3P({{E}_{2}})=0.6\] \[\Rightarrow \] \[0.4P({{E}_{2}})=0.1\] \[\Rightarrow \] \[P({{E}_{2}})=\frac{1}{4}\]
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