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Question: Two cards are drawn simultaneously from a well-shuffled deck of 52 cards. Find the probability distr...

Two cards are drawn simultaneously from a well-shuffled deck of 52 cards. Find the probability distribution of the number of successes, when getting a spade is considered a success.

Explanation

Solution

Hint: In this question it is given that two cards are drawn simultaneously from a well-shuffled deck of 52 cards. We have to find the probability distribution of the number of successes, when getting a spade is considered a success. So to find the solution we have to know that in a deck of 52 cards, there are 13 spades, and let X be the random variable denoting the number of success and success here is getting a spade for an event when two cards are drawn simultaneously. So we have to find the probability distribution for each random variable.
Complete step-by-step solution:
The number of successes is equal to some spades obtained in each draw. So the possible random variables are,
(i) X=0, i.e, selecting 0 spades.
(ii) X=1, i.e, selecting 1 spade.
(iii) X=2, i.e, selecting 2 spades.

(i) So when X=0, i.e, for selecting 0 spades, we removed all 13 spades from the deck and selected out of 39 cards.
P(X=0)=39C252C2P\left( X=0\right) =\dfrac{{}^{39}C_{2}}{{}^{52}C_{2}}
=(39!2!⋅(39−2)!)(52!2!⋅(52−2)!)\dfrac{\left( \dfrac{39!}{2!\cdot \left( 39-2\right) !} \right) }{\left( \dfrac{52!}{2!\cdot \left( 52-2\right) !} \right) }
=(39!2!⋅37!)(52!2!⋅50!)\dfrac{\left( \dfrac{39!}{2!\cdot 37!} \right) }{\left( \dfrac{52!}{2!\cdot 50!} \right) }
=39!⋅2!⋅50!2!⋅37!⋅52!\dfrac{39!\cdot 2!\cdot 50!}{2!\cdot 37!\cdot 52!}
=39×3852×51\dfrac{39\times 38}{52\times 51}=1934\dfrac{19}{34}.

(ii) now for X=1, i.e, for selecting 1 spade, we need to select 1 out of 13 spades and another one from 39 cards.
P(X=1)=13C1×39C152C2P\left( X=1\right) =\dfrac{{}^{13}C_{1}\times^{39} C_{1}}{{}^{52}C_{2}}
=13×39×252×51\dfrac{13\times 39\times 2}{52\times 51}
=1334\dfrac{13}{34}

(iii) for X=2, i,e, for selecting 2 spades, we need to select and 2 out of 13 spades,
P(X=2)=13C252C2P\left( X=2\right) =\dfrac{{}^{13}C_{2}}{{}^{52}C_{2}}
=13×12×22×52×51\dfrac{13\times 12\times 2}{2\times 52\times 51}
=234\dfrac{2}{34}
Therefore, the probability distribution PiP_{i} for each random variable XiX_{i},
P0=1934P_{0}=\dfrac{19}{34}, for X=0
P1=1334P_{1}=\dfrac{13}{34}, for X=1
P2=234P_{2}=\dfrac{2}{34}, for X=2
Note: So to solve this type of question you have to know that probability P=n(E)n(S)\dfrac{n\left( E\right) }{n\left( S\right) } , where n(E)= number of favourable outcome and n(S)= Total number of outcomes. Also nCr{}^{n}C_{r} defines choosing r number of quantity from n, which can be written as, nCr=n!r!⋅(n−r)!{}^{n}C_{r}=\dfrac{n!}{r!\cdot \left( n-r\right) !}.