Question

3 cards drawn simultaneously from pack of cards find the probability that one of them is red another spade and third is club

Answer 1

As per the data given in the above question.

We have to find the probability of one red,one spade and a club card .

Step -by-Step explanation:

Given ,

3 cards drawn simultaneously from a pack of card

Total number of cards in a pack = 52

STEP-1:

So,Let s sample space

3 card drawn from 52 cards = ways

$n(s) = C(52,3)$

$n(s) = \frac{52!}{(52 - 3)! \times 3!}$

$n(s) = \frac{52!}{(49)! \times 3!}$

Expand 52! upto 49!

$n(s) = \frac{52 \times 51 \times 50 \times 49!}{(49)! \times 3!}$

$n(s) = \frac{52 \times 51 \times 50}{ 3!}$

Now, Expand 3!

$n(s) = \frac{52 \times 51 \times 50}{ 3 \times 2 \times 1}$

$n(s) = \frac{132600}{ 6}$

$n(s) = 22100$

STEP-2:

Let n(T) be the event of getting three cards that is one red,one spade and a club card .

1. Number of choosing a red card from 26 - C(26,1)
2. Number of choosing a spade card from 13 - C(13,1)
3. Number of choosing a club card from 13- C(13,1)

$n(T)= \\ C(26,1)×C(13,1)×C(13,1)$

We know that C(26,1)= 26

Similarly, we acn write as

$n(T)= 26 \times 13 \times 13$

$n(T)= 4394$

STEP-3:

Let Probability be P(T)

$P(T) = \frac{n(T)}{n(s)}$

Put the values from above terms

$P(T) = \frac{4394}{22100}$

Hence ,the probability of getting one red,one spade and a club card simultaneously is 4394\22100.

Project code #SPJ2

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Answer 2

Answer:

Step-by-step explanation:

As per the data given in the above question.

We have to find the probability of one red, one spade and a club card .

Given ,

3 cards drawn simultaneously from a pack of card

Total number of cards in a pack = 52

STEP-1:

So, let s sample space

3 card drawn from 52 cards = ways

$n(s)= C (52,3)\\n(s)= \frac{52!}{(52-3)!\times3!}\\ n(s)=\frac{52!}{49!\times3!}$

expanding 52! upto 49!

n(s)= 52x 52x 50 x49!/ 49!x 3!

n(s)= 52x 51 x50/ 3!

now, expanding 3!

we get,

n(s)= 52x 51x 50/ 3x2x1

n(s)= 132600/6

n(s)= 22100

STEP 2:

let n(T) be the event of getting three cards that is one red and club card.

1. number of choosing a red card from 25- C(26, 1)

2. number of choosing a spade card from 12- C(13, 1)

3. number of choosing a club card from 13-C(13,1)

n(T)= C(26,1)AC(13, 1) AC(13,1)

we know that C(26,1)=26

Similarly we can write as

n(s)= 26x 13x13

n(s)= 4394

let the probability be P(T)

p(T)= n(T)/ n(s)

putting the values

p(T)= 4394/ 22100

Hence the probability of getting one red , one spade and a club is 4394/ 22100