Math, asked by solankikshitij82, 3 months ago

from apack of 52plauing cards Jack King or queen King and ace of red colour are removed from the remaining a card drawn at randomfi d probablit​

Answers

Answered by muskanjangde861
0

Step-by-step explanation:

Total no. of cards in a pack =52

After removing red-colored - jack, queen, king, and aces,

No. of cards =52−8=44 and

No. of red cards =26−8=18.

Solution(i):

No. of black queens =2

Therefore,

2

C

1

( Selecting 1 out of 2 items) times out of

44

C

1

( Selecting 1 out of 44 items) a black queen is picked.

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

We know that, Probability P(E) =

(Total no.of possible outcomes)

(No.of favorable outcomes)

=

44

C

1

2

C

1

=

44

2

=

22

1

Solution(ii):

No. of red cards =18

Therefore,

18

C

1

( Selecting 1 out of 18 items) times out of

44

C

1

( Selecting 1 out of 44 items) a red card is picked.

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

We know that, Probability P(E) =

(Total no.of possible outcomes)

(No.of favorable outcomes)

=

44

C

1

18

C

1

=

44

18

=

22

9

Solution(iii):

No. of black jack cards =2

Therefore,

2

C

1

( Selecting 1 out of 2 items) times out of

44

C

1

( Selecting 1 out of 44 items) a black jack card is picked.

Let E be the event of getting a black jack card from the pack

We know that, Probability P(E) =

(Total no.of possible outcomes)

(No.of favorable outcomes)

=

44

C

1

2

C

1

=

44

2

=

22

1

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