Math, asked by akhilrana410, 1 day ago

Two cards are drawn at random from a pack of 52 cards. Then find the probability of getting one red face card and one black ace?

Answers

Answered by AkashAkshitha
0

Answer:

3/676

Step-by-step explanation:

1) Since, the two cards are drawn simultaneously, the outcome of first will not affect outcome of second.

2) Case 1 : red face card

Total red face cards in deck are :- 6

Total possible cards in the deck are :- 52.

Probability of getting a red face card would be  = 6/52 = 3/26

3) Case 2 : One black ace

Total black ace in the deck are : 2

Total possible cards in the deck are :- 52.

Probability of getting a black ace card would be  = 2/52 = 1/26

4) Overall probability would be : 3/26 * 1/26 = 3/ 676

Answered by amitnrw
1

Probability of getting one red face card and one black ace is 2 /221

≈ 0.009

Given : Two cards are drawn at random from a pack of 52 cards.

To find :  the probability of getting one red face card and one black ace

Solution:

Understand first , what all cards are there in a pack of 52 cards :

Four types : Club , Spade , Diamond & Heart

(Clubs & Spade are black in color. Diamond & heart are Red in color)

Each type has 13 cards :  

Ace – (number 1)  

Numbered card  2-10

Face Card – Jack (11) , Queen (12)  King (13)

2 Cards out of 52 can be drawn in ⁵²C₂  ways

red face card = 6

one red face card can be selected in 6 ways

Black ace = 2

one black ace can be selected in 2 ways

Probability of an event  = n(E)/n(S)  

n(E) = number of possible outcome of event

n(S) = number of possible sample space outcome

Probability of getting one red face card and one black ace

= ( 6 * 2) / ⁵²C₂

= 12 * 2 / (52 * 51)

= 2/ ( 13 * 17)

= 2 /221

≈ 0.009

Probability of getting one red face card and one black ace is 2 /221

≈ 0.009

Learn More:

the probability that an event happens in one trial of an experiment is ...

brainly.in/question/8759895

Determine the experimental probability of getting a head (or a tail ...

brainly.in/question/18400413

Similar questions