A) \[\frac{1}{10}\]
B) \[\frac{3}{10}\]
C) \[\frac{3}{5}\]
D) \[\frac{1}{2}\]
Correct Answer: D
Solution :
| [d] Let \[{{A}_{i}}(i=2,3,4,5)\] be the event that urn contains \[2,3,4,5\] white balls and let B be the event that two white balls have been drawn than we have to find \[P({{A}_{5}}/B).\] |
| Since the four events \[{{A}_{2}},{{A}_{3}},{{A}_{4}}\] and \[{{A}_{5}}\] are equally likely we have \[P({{A}_{2}})=P({{A}_{3}})=P({{A}_{4}})\] |
| \[=P({{A}_{5}})=\frac{1}{4}.\] |
| \[P(B/{{A}_{2}})\] is probability of event that the urn contains 2 white balls and both have been drawn. |
| \[\therefore P(B/{{A}_{2}})=\frac{^{2}{{C}_{2}}}{^{5}{{C}_{2}}}=\frac{1}{10}\] |
| Similarly \[P(B/{{A}_{3}})=\frac{^{3}{{C}_{2}}}{^{5}{{C}_{2}}}=\frac{3}{10},\] |
| \[P(B/{{A}_{4}})=\frac{^{4}{{C}_{2}}}{^{5}{{C}_{2}}}=\frac{3}{5},P(B/{{A}_{5}})=\frac{^{5}{{C}_{2}}}{^{5}{{C}_{2}}}=1.\] |
| By Baye?s theorem, |
| \[P({{A}_{5}}/B)=\frac{P({{A}_{5}})P(B/{{A}_{5}})}{(P({{A}_{2}})P(B/{{A}_{2}})+P({{A}_{3}})P(B/{{A}_{3}})}\] |
| \[+P({{A}_{4}})(B/{{A}_{4}})+P({{A}_{5}})P(B/{{A}_{5}}))\] |
| \[=\frac{\frac{1}{4}.1}{\frac{1}{4}\left[ \frac{1}{10}+\frac{3}{10}+\frac{3}{5}+1 \right]}=\frac{10}{20}=\frac{1}{2}.\] |
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