Question

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 ?

Answer 1

red face cards are 6

black faces are 2

required probability

$= \frac{6 \times 2}{ \binom{52}{2} }$

$\frac{12}{ \frac{52 \times 51}{2} }$

$\frac{12 \times 2}{52 \times 51}$

$\frac{2}{17 \times 13}$

$= \frac{2}{221}$

Answer 2

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

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