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# daily question :) (hard)

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an experiment consists of picking a card from a standard deck of playing cards and drawing a counter from a bag that contains 5 counters: 2 blue, 2 white, 1 red.

A) picking a spade and drawing a blue counter

B) picking a red card and drawing a red counter

C) picking a face card and not drawing a white counter

D) picking a diamond and drawing a green counter

cheifraidhunter  Jun 6, 2017
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#1
+85821
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A) picking a spade and drawing a blue counter

P(spade)  =  1/4   ...  P(blue) =  2/5

So    P(spade) and P(blue)  =    1/4 * 2/5  =   2/20   =  1/10

B) picking a red card and drawing a red counter

P(red card)  = 1/2       .... P(red counter)  = 1/5

So....  P(red card) and P(red counter)  =   1/2 * 1/5  =   1/10

C) picking a face card and not drawing a white counter

P(face card)  = 12/52  =  3/13  .....  P(not white)  = 3/5

So..   P(face card) * P(not white)   =   3/13 * 3/5   =  9 / 65

D) picking a diamond and drawing a green counter

P(diamond)  = 1/4     ...   P(green)  = 0/5  = 0

So   P(diamond) * P(green)   =   (1/4) * 0   =  0

CPhill  Jun 6, 2017
#2
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A: Spades are one of the 4 suits in the 52 card deck that is standard

13/52

Multiply by the blue counters over the total counters.

13/52 * 2/5 = 26/260 OR 1/10

1/10 chance of choosing a spade and a blue counter.

B: Red cards are half of the deck

1/2

Multiply by red counter over total

1/2*1/5= 1/10

1/10 chance of choosing a red card and red counter

C: Face cards are Jack, Queen, King. There are 12 in a standard deck

12/52

NOT choosing white means taking blue and red over the total counters

3/5

Multiply

12/52 * 3/5 = 36/260 OR 9/65

9/65 chance of choosing a face card and not choosing a white counter

D: Diamonds make up 1/4 of the deck

1/4

There are no green counters

Probability of choosing a diamond AND a green counter is 0.

rarinstraw1195  Jun 6, 2017

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