A man throws a die until he gets a number greater than `3`. The probability that he gets `5` in the last throw
A. `1//3`
B. `1//4`
C. `1//6`
D. `1//36`
Correct Answer - A
`(a)` At event `A` he gets a number greater than `3`
`:.P(A)=1//2`
At event `B` he gets `5` in last throw
`:.P(B)=1//6`
`:.` Required probability is
`P(E)=(1)/(6)+(1)/(2)*(1)/(6)+(1)/(2)*(1)/(2)*(1)/(6)+…..oo`
`=((1)/(6))/(1-(1)/(2))=(1)/(3)`
Correct Answer - C
Given that 5 and 6 have appeared on two of the dice, the sample space reduces to `6^(4) - 2 xx 5^(4) + 4^(4)` (inclusion-exclusion principle). Also,...
Correct Answer - D
For `Xge6,` the probability is
`(5^(5))/(6^(6))+(5^(5))/(6^(7))+...oo=(5^(5))/(6^(6))((1)/(1-5//6))=((5)/(6))^(5)`
For `Xgt3,`
`(5^(3))/(6^(4))+(5^(4))/(6^(5))+(5^(5))/(6^(6))+...oo=((5)/(6))^(3)`
Hence, the conditional probability is
`((5//6)^(6))/((5//6)^(3))=25/36`