Question

One card is drawn at random from a well-shuffled deck of $52$ cards. In how many of the following cases are the events $E$ and $F$ independent?
(i) $E:$ 'the card drawn is a spade'
$F:$ 'the card is drawn is an ace'
(ii) $E:$ 'the card drawn is black'
$F:$ 'the card drawn is a king'
(iii) $E:$ '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.
$\therefore P\left(E\right)=P\left(thecarddrawnisaspade\right)=\frac{13}{52}=\frac{1}{4}$
$\therefore P\left(F\right)=P\left(thecarddrawnisanace\right)=\frac{4}{52}=\frac{1}{13}$
In the deck of cards, only $1$ card is an ace of spades.
$P\left(EF\right)=P\left(thecarddrawnisspadeandanace\right)=\frac{1}{52}$
$P\left(E\right)\times P\left(F\right)=\frac{1}{4}\cdot \frac{1}{13}=\frac{1}{52}=P\left(EF\right)$
$\Rightarrow P\left(E\right)\times P\left(F\right)=P\left(EF\right)$
Therefore, the events $E$ and $F$ are independent.
(ii) In a deck of $52$ cards, $26$ cards are black and $4$ cards are kings.
$\therefore P\left(E\right)=P\left(thecarddrawnisblack\right)=\frac{26}{52}=\frac{1}{2}$
$\therefore P\left(F\right)=P\left(thecarddrawnisaking\right)=\frac{4}{52}=\frac{1}{13}$
In the pack of $52$ cards, $2$ cards are black as well as kings.
$\therefore P\left(EF\right)=P\left(thecarddrawnisablackking\right)=\frac{2}{52}=\frac{1}{26}$
$P\left(E\right)\times P\left(F\right)=\frac{1}{2}\cdot \frac{1}{13}=\frac{1}{26}=P\left(EF\right)$
Therefore, the given events $E$ and $F$ are independent.
(iii) In a deck of $52$ cards, $4$ cards are kings, $4$ cards are queens, and $4$ cards are jacks.
$\therefore P\left(E\right)=P\left(thecarddrawnisakingoraqueen\right)=\frac{8}{52}=\frac{2}{13}$
$\therefore P\left(F\right)=P\left(thecarddrawnisaqueenorajack\right)=\frac{8}{52}=\frac{2}{13}$
There are $4$ cards which are king or queen and queen or jack.
$\therefore P\left(EF\right)=P(thecarddrawnisakingoraqueen,orqueenorajack)=\frac{4}{52}=\frac{1}{13}$
$P\left(E\right)\times P\left(F\right)=\frac{2}{13}\cdot \frac{2}{13}=\frac{4}{169}\ne \frac{1}{13}$
$\Rightarrow P\left(E\right)\cdot P\left(F\right)\ne P\left(EF\right)$
Therefore, the given events $E$ and $F$ are not independent.