Question

4 cards are drawn. Probability of getting one diamond and 3 spade

Answer 1

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Answer 2

   

I've a doubt on drawing the 4 cards either one by one or together at one time. I'm assuming that the 4 cards are drawn at one time.

We know that there are 52 cards, of which 13 are spades, 13 are diamonds, 13 are clubs and 13 are hearts. If we want to choose 4 cards at random from these, the total no. of outcomes of selecting 4 cards from these 52 cards is '52C4', which values:

$\Rightarrow\ \ ^{52}C_4 \\ \\ \Rightarrow\ \frac{52!}{4! \times 48!} \\ \\ \\ \Rightarrow\ \frac{52 \times 51 \times 50 \times 49}{24} \\ \\ \\ \Rightarrow\ 13 \times 17 \times 25 \times 49 \\ \\ \Rightarrow\ 270725$

So that we can select 4 random from the total of 52 cards in 270725 ways. So the total no. of outcomes is 270725. Now we've to find in how many ways we can select 1 diamond and 3 spade.

Among these 270725 outcomes, there are many '1 diamond and 3 spades' possibilities. I'm letting them as a set. For this set, we can select 1 diamond from the total 13 diamonds in 13C1 = 13 ways and 3 spades from the total 13 spades in 13C3 = 286 ways. So we can take 1 diamond and 3 spades in 13 × 286 = 3718 ways, so this set contains 3718 possibilities. This is the no. of favourable outcomes.

∴ Probability,

$\Rightarrow\ \frac{3718}{270725} \\ \\ \Rightarrow\ \bold{\frac{286}{20825}}$

So the probability is 286/20825.

Thank you...