Face : | 1 | 2 | 3 | 4 | 5 | 6 |
Probability : | 0.2 | 0.22 | 0.11 | 0.25 | 0.05 | 0.17 |
A) \[\frac{1}{6}\]
B) \[\frac{1}{4}\]
C) \[\frac{5}{6}\]
D) None of these
Correct Answer: C
Solution :
Let \[A\] be the event that face 4 turns up and \[B\] be the event that face 5 turns up then \[P(A)=0.25,\] \[P(B)=0.05\]. Since \[A\] and \[B\] are mutually exclusive, so \[P(A\cup B)=P(A)+P(B)=0.25+0.05=0.30\]. We have to find \[P\left( \frac{A}{A\cup B} \right),\] which is equal to \[P\frac{[A\cap (A\cup B)]}{P(A\cup B)}=\frac{P(A)}{P(A\cup B)}=\frac{0.25}{0.30}=\frac{5}{6}\].You need to login to perform this action.
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