Question

Probability fo solving specific problem independently by A and B are $\frac{1}{2}and\frac{1}{3}$ respectively. If both try to solve the problem independently, Find the probability that
the problem is solved

Probability fo solving specific problem independently by A and B are $\frac{1}{2}and\frac{1}{3}$ respectively. If both try to solve the problem independently, find the probability that
Exactly one of them solves the problem

One card is drawn at random from a well- shuffled deck of 52 cards. In which of the following cases are the events E and F independent?
E: the card drawn is a spade, F: the card drawn is an ace

Answer 1

Probability fo solving the problem by $A,P\left(A\right)=\frac{1}{2}$

Probability of solving the problem by $B,P,\left(B\right)=\frac{1}{3}$
Probability of not solving the problem by $A=P\left(A\right)=1-P\left(A\right)=1-\frac{1}{2}=\frac{1}{2}$
and probability of not solving the problem by B
$=P\left(B\right)=1-P\left(B\right)=1-\frac{1}{3}=\frac{2}{3}$
P (the problem is solved) = 1- P (none of them solve the problem)
$=1-P(A^{\prime }\cap B^{\prime })=1-P\left(A\right)P\left(B\right)$
$(\therefore AandBareindependent\Rightarrow A^{\prime }and{B}^{\prime }areindependent)$
$=1-(\frac{1}{2}\times \frac{2}{3})=1-\frac{1}{3}=\frac{2}{3}$

Probability fo solving the problem by $A,P\left(A\right)=\frac{1}{2}$

Probability of solving the problem by $B,P,\left(B\right)=\frac{1}{3}$
Probability of not solving the problem by $A=P\left(A^{\prime }\right)=1-P\left(A\right)=1-\frac{1}{2}=\frac{1}{2}$
and probability of not solving the problem by B
$=P\left(B^{\prime }\right)=1-P\left(B\right)=1-\frac{1}{3}=\frac{2}{3}$
P (Exactly one of them solve the problem) = P (A) P(B') + P(A') P(B)
$=\frac{1}{2}\times \frac{2}{3}+\frac{1}{2}\times \frac{1}{3}=\frac{1}{3}+\frac{1}{6}=\frac{2+1}{6}=\frac{3}{6}=\frac{1}{2}$

In a deck of 52 cards , 13 cards are spades and 4 cards are aces.
Given, E: the card drawn is a spade $\Rightarrow n\left(E\right)=13,$
and F: the card drawn is an ace
$\Rightarrow n\left(F\right)=4andn\left(S\right)=52$
Here, P (E) = P (card drawn isspade) $=\frac{n\left(E\right)}{n\left(S\right)}=\frac{13}{52}=\frac{1}{4}$
P(F) = P (card drawn is an ace) $=\frac{4}{52}=\frac{1}{13}$
Also, E $\cap$ F: the deck of cards, only 1 card is an ace of spades,
$\Rightarrow P(E\cap F)=\frac{n(E\cap f)}{n\left(S\right)}=\frac{1}{52}$
Now, $P\left(E\right)\times P\left(F\right)=\frac{1}{4}\times \frac{1}{13}=\frac{1}{52}=P(E\cap F)$