Question #187737

Picking at the same time 6 cards from a standard deck of playing.

Find the following probability:

a)  At least 2 king cards

b)  None of the cards ace

c)   At most 4 black cards

d)  All cards are red

this is the homework please email me the answers back asap

thanks.


1
Expert's answer
2021-05-07T11:15:02-0400

The number of all possible outcomes is C(52,6)C(52,6) because we take 6 cards from 52 cards deck.

a) we want to have 2, 3, or 4 kings and any other cards in the rest of 6 cards.

P(A)=C(4,2)C(50,4)C(52,6)+C(4,3)C(49,3)C(52,6)+C(4,4)C(48,2)C(52,6)=6230300+418424+1112820358520=145662420358520=0.0715\displaystyle P(A) = \frac{C(4,2) \cdot C(50,4)}{C(52,6)} + \frac{C(4,3) \cdot C(49,3)}{C(52,6)} + \frac{C(4,4) \cdot C(48,2)}{C(52,6)} = \frac{6 \cdot 230300 + 4 \cdot 18424 + 1 \cdot 1128 }{20358520} = \frac{1456624}{20358520} =0.0715

b) none are aces means that we can select any from 48 cards (52 - 4 aces)

P(B)=C(48,6)C(52,6)=1227151220358520=0.603\displaystyle P(B) = \frac{C(48, 6)}{C(52,6)} = \frac{12271512}{20358520} = 0.603

c) P(C) = 1 - P(5 or 6 black cards)

P(C)=1C(26,5)C(26,1)C(52,6)C(26,6)C(52,6)=10.0840.011=0.905\displaystyle P(C) = 1 - \frac{C(26,5) \cdot C(26,1)}{C(52,6)} - \frac{C(26,6) }{C(52,6)} = 1 - 0.084 - 0.011= 0.905

d) P(D)=C(26,6)C(52,6)=0.0113\displaystyle P(D) = \frac{C(26,6)}{C(52,6)} = 0.0113


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