A stack of ten cards has four hearts (which are red), three diamonds (which are

Question

A stack of ten cards has four hearts (which are red), three diamonds (which are red), two spades (which are black) and one club (which is black). Four cards are sampled from the stack, randomly and without replacement, and placed in a box. Then one card is randomly selected from the box. If that card is a heart then all of the other three cards in the box are burnt. Otherwise, two cards are sampled, randomly and without replacement, from the other three cards in the box, and then these two cards are burnt. Find the probability that exactly k red cards are burnt, for each possible value of k

Explanation / Answer

Stack of ten cards has four hearts, three diamonds, two spades and one club. So, there are 7 red cards and three black.

So here in first scenario, if first card is heart, then we will burnt other three cards.

so in that case k can have values 0,1,2 or 3

Now, as we know that first card is heart, in the case 1, so that leaves us with 9 card, out of which 6 remaining are red.

so Here

p(k' = 0) = p(no red card in remaining three cards l first one is a heart) = 6C0 * 3C3/ 9C3 = 0.0119

p(k' = 1) =  p(no red card in remaining three cards l first one is a heart) = 6C1 * 6C2/ 9C3 = 0.2143

p(k' = 2) =  p(no red card in remaining three cards l first one is a heart) = 6C2 * 6C1/ 9C3 = 0.5357

p(k' = 3) =  p(no red card in remaining three cards l first one is a heart) = 6C3 * 6C0/ 9C3 = 0.2381

so now that was the first case where first drawn card out of four was Heart so Pr(First drawn card is heart) = 0.4

Now , if it was not heart for the first card than we will draw two more card that mean now k can have three value k = 0,1 and 2 as there can be 0,1 or 2 cards out of the next two cards. Here in this case there are two more cases

(i) When the first drawn card is diamonds (red) so number of remaining red cards are 6

(ii) when the first drawn card is black so number of remaining red cards are 7

so for first case

p(k''= 0) = p(no red card in selcted two cards out of three l first one is a non-heart and a diamond) = 6C0 * 3C2/ 9C2 = 0.0833

p(k'' = 1) =  p(no red card in selcted two cards out of three l first one is a non-heartand a diamond) = 6C1 * 3C1/ 9C2 = 0.5

p(k'' = 2) =  p(no  red card in selcted two cards out of three l first one is a non-heartand a diamond) = 6C2 * 3C1/ 9C2 = 0.4167

Here Pr(First drawn card to be diamond when first card is not heart ) = 0.3/0.6 = 0.5

Now the option (2) of case(ii) where first drawn card is black so there are seven red cards are here

p(k'''= 0) = p(no red card in selcted two cards out of three l first one is a non-heart and a black) = 7C0 * 2C2/ 9C2 = 0.0278

p(k''' = 1) =  p(no  red card in selcted two cards out of three l first one is a non-heart and a black) = 7C1 * 2C1/ 9C2 = 0.3889

p(k''' = 2) =  p(no  red card in selcted two cards out of three l first one is a non-heart and a black)  = 7C2 * 2C0/ 9C2 = 0.5833

Pr(first drawn card is a black when first drawn card in nonheart) = 0.3/0.6 = 0.5

So now we will add up all this probabilityies

so for final probability distribution of k

p(k) = p(k') * 0.4 + p(k'') * 0.6 * 0.5 + P(k''') * 0.6 * 0.5 = p(k') * 0.4 + p(k'') * 0.3 + P(k''') * 0.3

p(k =0) = 0.4 * 0.0119 + 0.3 * (0.0833 + 0.0278)) = 0.0381

p(k = 1) = 0.4 * 0.2143 + 0.3 * (0.5 + 0.3889) = 0.3524

p(k = 2) = 0.4 * 0.5357 + 0.3 * (0.4167 + 0.5833) = 0.5143

p(k = 3) = 0.4 * 0.2381 + 0.3 * (0 + 0) = 0.0952

so,

p(k) = 0.0381 ; k = 0

= 0.3524 ; k = 1

= 0.5143 ; k = 2

= 0.0952 ; k = 3

E(k) = 5/3

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.