Question

hi friends plz answer..............

Answer 1

hey mate here is the answer of imo question

red face cards=6

n(sample space)=52-6=46

now

a)let E1 be the event that the remove card is a face card

so n(E1)=12-6=6.......as 6 red face cards are removed

so p(E1)=n(E1)/n(S)

=6/46

=3/23

b)let E2 be the event that the remove card is a king card

so n(E2)=4-2=2.........as 2 red king cards are removed

so p(E2)=n(E2)/n(S)

=2/46

=1/23

hence option C is correct

hope it will help you

mark me brainliest ✌✌✌✌

Answer 2

Total number of face card in pack of card = 12

In which there are 6 red cards and 6 black cards.

>>> (I) <<<

So, probability for face card after removing all red face cards,

$p1 = \frac{6}{52 - 6}$

$p1 = \frac{6}{46}$

$p1 = \frac{3}{23}$
Option (C)

>>> (II) <<<

There are total 4 king cards is pack of card

In which 2 are red and 2 are black cards,

Therefore probability for king card is,

$p2 = \frac{2}{52 - 6}$

$p2 = \frac{2}{46}$

$p2 = \frac{1}{23}$
Option (C)