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Question

Two cards are drawn together from a pack of 52 cards. The probability that one is a club and one is a diamond?
Options:
A .  $MF#%\dfrac{13}{51}$MF#%
B .  $MF#%\dfrac{1}{52}$MF#%
C .  $MF#%\dfrac{13}{102}$MF#%
D .  $MF#%\dfrac{1}{26}$MF#%
Answer: Option C

Answer : Option C

Explanation :

n(S) = Total number of ways of drawing 2 cards from 52 cards = 52C2
Let E = event of getting 1 club and 1 diamond.
We know that there are 13 clubs and 13 diamonds in the total 52 cards.
Hence, n(E) = Number of ways of drawing one club from 13 and one diamond from 13
= 13C1 × 13C1

$MF#%\text{P(E) = }\dfrac{\text{n(E)}}{\text{n(S)}} = \dfrac{13_{C_1} \times 13_{C_1}}{52_{C_2}}$MF#%

$MF#%= \dfrac{13 \times 13}{\left( \dfrac{52 \times 51}{2}\right)}
= \dfrac{13 \times 13}{ 26 \times 51}= \dfrac{13}{ 2\times 51}= \dfrac{13}{102}
$MF#%



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