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.


Expert's answer

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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