A card from a pack of 52 cards is lost. From | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $E_{1}$ be the event that the lost card is a diamond and $E_{2}$ be the event that the lost card is not a diamond.Let $A$ be the event that tw

Vedclass · Accepted Answer

$\frac{11}{50}$

Vedclass · Answer

(A) Let $E_{1}$ be the event that the lost card is a diamond and $E_{2}$ be the event that the lost card is not a diamond.Let $A$ be the event that two cards drawn from the remaining $51$ cards are both diamonds.Out of $52$ cards,$13$ are diamonds and $39$ are not diamonds.$P(E_{1}) = \frac{13}{52} = \frac{1}{4}$ and $P(E_{2}) = \frac{39}{52} = \frac{3}{4}$.If $E_{1}$ occurs,there are $12$ diamonds left in $51$ cards. The probability of drawing $2$ diamonds is $P(A|E_{1}) = \frac{^{12}C_{2}}{^{51}C_{2}} = \frac{12 	imes 11}{51 	imes 50} = \frac{132}{2550}$.If $E_{2}$ occurs,there are $13$ diamonds left in $51$ cards. The probability of drawing $2$ diamonds is $P(A|E_{2}) = \frac{^{13}C_{2}}{^{51}C_{2}} = \frac{13 	imes 12}{51 	imes 50} = \frac{156}{2550}$.Using Bayes' theorem,$P(E_{1}|A) = \frac{P(E_{1})P(A|E_{1})}{P(E_{1})P(A|E_{1}) + P(E_{2})P(A|E_{2})}$.$P(E_{1}|A) = \frac{\frac{1}{4} 	imes \frac{132}{2550}}{\frac{1}{4} 	imes \frac{132}{2550} + \frac{3}{4} 	imes \frac{156}{2550}} = \frac{132}{132 + 3 	imes 156} = \frac{132}{132 + 468} = \frac{132}{600}$.Dividing by $12$,we get $\frac{11}{50}$.

$A$ card from a pack of $52$ cards is lost. From the remaining cards of the pack,two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.

Similar Questions

There are two balls in an urn. Each ball can be either white or black. If a white ball is put into the urn and thereafter a ball is drawn at random from the urn,then the probability that it is white is

$25 \%$ of the population are smokers. $A$ smoker has $27$ times more chances to develop lung cancer than a non-smoker. $A$ person is diagnosed with lung cancer and the probability that this person is a smoker is $\frac{k}{10}$. Then the value of $k$ is $.............$

$A$ pack of playing cards was found to contain only $51$ cards. If the first $13$ cards which are examined are all red,then the probability that the missing card is black,is

$A$ bag contains four balls. Two balls are drawn randomly and found to be white. The probability that all the balls in the bag are white is

Vedclass Test Series

Exam Paper Generator

Online Exam Module