Probability question

[alexamasan]

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If three cards are selected from an ordinary deck, find the probability that all 3 are black.

I can't seem to understand how to solve this problem.

 

[Valigator00]

If three cards are selected from an ordinary deck, find the probability that all 3 are black.

I can't seem to understand how to solve this problem.

Deck has 52 cards. 26 are red, 26 are black. Probability of picking first black card is 26/52 or 1/2. Probability of picking another black card is 25/51. Probability of picking a third black card is 24/50.
So:
(1/2)x(25/51)x(24/50)=(1/1)x(1/51)x(6/1)
I did this by reducing the fractions.

Multiply this across to get 6/51.

 

[whawha]

Deck has 52 cards. 26 are red, 26 are black. Probability of picking first black card is 26/52 or .5. Probability of picking another black card is 25/51. Probability of picking a third black card is 24/50. Multiply this across and simplify to get answer. I believe it is 6/51

Yup, and in the problem, the fact that cards are drawn without replacement has to be stated 🙂 Otherwise, it would have been (26/52)*(26/52)*(26/52) 🙂

 

[Valigator00]

Yup, and in the problem, the fact that cards are drawn without replacement has to be stated 🙂 Otherwise, it would have been (26/52)*(26/52)*(26/52) 🙂

Correct. If they were drawn with replacement, problem would be even more simple. (Answer would be 1/8)

 

[alexamasan]

Thanks a lot, that's what I tried to do, but it seems I made a mistake in the arithmetic😳. Thanks again.

 

[S2000]

(1/2)x(25/51)x(24/50)=(1/1)x(1/51)x(6/1)
I did this by reducing the fractions.

How did your reduce 25/51 to 1/51 and 24/50 to 6/1? I must be misunderstanding. Sorry if thats a dumb question.

 

[UCB05]

How did your reduce 25/51 to 1/51 and 24/50 to 6/1? I must be misunderstanding. Sorry if thats a dumb question.

He's just cancelling out factors.
(1/2)x(25/51)x(24/50)=
(1/1)x(25/51)x(12/50)=
(1/1)x(1/51)x(12/2)=
(1/1)x(1/51)x(6/1)

 

[kkmaths90]

so if it doesn't say without replacement then that means that put the card back right?