\[x\] | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
\[p(x)\] | 0.15 | 0.23 | 0.12 | 0.10 | 0.20 | 0.08 | 0.07 | 0.05 |
A) 0.87
B) 0.77
C) 0.35
D) 0.50
Correct Answer: B
Solution :
If A and B are two events associated with a random experiment. Then, \[P(A\cup B)=P(A)+P(B)-P(A\cap B),\] which is equal to the atleast one of the event is occur. E = {X is a prime number} = { 2, 3, 5, 7} \[P(E)=P(X=2)+P(X=3)\] \[+P(X=5)+P(X=7)\] \[\Rightarrow \]\[P(E)=0.23+0.12+0.20+0.07+0.62\] and \[F=\{X<4\}=\{1,2,3\}\] \[P(F)=P(X=1)+P(X=2)+P(X=3)\] \[\Rightarrow \] \[P(F)=0.15+0.23+0.12=0.5\] \[E\cap F=\]{X is prime number as well as < 4} = {2, 3} \[P(E\cap F)=P(X=2)+P(X=3)\] \[=0.23+0.12=0.35\] \[\therefore \]Required probability, \[P(E\cup F)=P(E)+P(F)-P(E\cap F)\] \[P(E\cup F)=0.62+0.5-0.35\] \[P(E\cup F)=0.77\]You need to login to perform this action.
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