Question

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

E : the card drawn is black, F: the card drawn is a king

E: the card drawn is a king or queen
F: the card drawn is a queen or jack

Answer 1

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, $P\left(E\right)\times P\left(F\right)=\frac{1}{4}\times \frac{1}{13}=\frac{1}{52}=P(E\cap F)$

In a deck of 52 cards, 26 cards are black and 4 cards are kings. Given, E : the cards drawn is black $\Rightarrow n\left(E\right)=26$
F: the cards drawn is kng $\Rightarrow n\left(F\right)=4$
Also, n(S) = 52
P(E) = P (cards drawn is black) = $\frac{n\left(E\right)}{n\left(S\right)}=\frac{26}{52}=\frac{1}{2}$
and P(F) = P (card drawn is a king) $\frac{n\left(E\right)}{n\left(S\right)}=\frac{4}{52}=\frac{1}{13}$
Also, $E\cup F:$ card drawn is a black king
$\Rightarrow P(E\cap F)=\frac{n(E\cap S)}{n\left(S\right)}=\frac{2}{52}=\frac{1}{26}Now,p\left(E\right)\times P\left(F\right)=\frac{1}{2}\times \frac{1}{13}=\frac{1}{26}P(E\cap F)\Rightarrow P(E\cap F)=P\left(E\right)P\left(F\right)$

In a deck of 52 cards, 4 cards are kings, 4 cards are queens and 4 cards are jacks.
P(E) = P ( card drawn is a king or a queen)
= P (King)+ P (Queen) = $\frac{4}{52}+\frac{4}{52}=\frac{8}{52}=\frac{2}{13}$
and P (F) = P (cards drawn is a queen or a jack)
$=P\left(Queen\right)+\left(Jack\right)=\frac{4}{52}+\frac{4}{52}=\frac{8}{52}=\frac{2}{13}$
Now, $P\left(E\right)\times P\left(F\right)=\frac{2}{13}\times \frac{2}{13}=\frac{4}{169}\ne P(E\cap F)\Rightarrow P(E\cap F)\ne P\left(E\right)P\left(F\right)$
Therefore, the events E and F are not independent.