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A dice is thrown. The probability that the first time 1 occurs at the even throw isA. `1//6`B. `5//11`C. `6//11`D. `5//36` |
Answer» Correct Answer - B Let E be the event of getting 1 on a die . `implies P(E ) =(1)/(6) and P(bar (E )) =(5)/(6)` `:.` P (first time 1 occurs at the even throw ) `=t_(2) or t_(4) or t_(6) or t(8) …… and ` so on `={P(bar( E)) P(E ) }+{P(bar( E))P(bar(E )) P(bar(E )) }+ ... oo ` `((5)/(6))((1)/(6))+((5)/(3))^(3)((1)/(6))+((5)/(6))^(5)((1)/(6))+oo=((5)/(36))/(1-(25)/(36))=(5)/(11) ` |
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