Question

Red queens and black jacks are removed from a pack of 52 playing cards. Find the probability that the card drawn from the remaining cards is: (i) a card of clubs or an ace (ii) a black king (iv) either a king or a queen (iii) neither a jack nor a king​

Answer 1

Answer:

Total no. of cards in pack =52

After removing red queens(2), and black jacks(2)

No. of cards =52−4=48

Solution(i):

No. of kings =4

Therefore,

4

C

1

( Selecting 1 out of 4 items) times out of

48

C

1

( Selecting 1 out of 48 items) a king is picked.

Let E be the event of getting a king from pack

We know that, Probability P(E) =

(Total no.of possible outcomes)

(No.of favorable outcomes)

=

48

C

1

4

C

1

=

48

4

=

12

1

Solution(ii):

No. of red colors =26−2=24......(2 red queen cards already drawn)

Therefore,

24

C

1

( Selecting 1 out of 24 items) times out of

48

C

1

( Selecting 1 out of 48 items) a red is picked.

Let E be the event of getting red from the pack

We know that, Probability P(E) =

(Total no.of possible outcomes)

(No.of favorable outcomes)

=

48

C

1

24

C

1

=

48

24

=

2

1

Solution(iii):

No. of face cards =8 ....... (2 red queen, 2 black jack (face cards) already drawn)

Therefore,

8

C

1

( Selecting 1 out of 8 items) times out of

48

C

1

( Selecting 1 out of 48 items) a face card is picked.

Let E be the event of getting a face card from the pack

We know that, Probability P(E) =

(Total no.of possible outcomes)

(No.of favorable outcomes)

=

48

C

1

8

C

1

=

6

1

Solution(iv):

No. of queen cards =4−2=2...(2 red queen cards already drawn)

Therefore,

2

C

1

( Selecting 1 out of 2 items) times out of

48

C

1

( Selecting 1 out of 48 items) a queen is picked.

Let E be the event of getting a queen from the pack

We know that, Probability P(E) =

(Total no.of possible outcomes)

(No.of favorable outcomes)

=

48

C

1

2

C

1

=

24

1

thankyou