Math Is Fun Forum

mathxyz

Member

From: Brooklyn, NY

Registered: 2024-02-24

Posts: 1,053

Jar of Sweets

A jar consists of 21 sweets. 12 are green and 9 are blue. William picked two sweets at random.

Find the probability that

1) both sweets are blue.

2) one sweet is blue and one sweet is green.

Let me see.

Part 1

Let A = probability that first sweet is blue

Let B = probability that the second sweet is also blue

P(A n B) = P(A) • P(A|B)

P(A) = 9/21

P(A|B) = 8/20

P(A n B) = (9/21) • (8/20)

Is this correct?

Part 2

A = probability sweet is blue.

B = probability sweet is green.

P(A n B) = P(A) • P(B)

P(A) = 9/21

P(B) = 12/21

P(A n B) = (9/21) • (12/21)

Is this correct?

Keep_Relentless

Member

From: Queensland, Australia

Registered: 2024-05-05

Posts: 61

Re: Jar of Sweets

For part 2, I would do (9/21)*(12/20) + (12/21)*(9/20). That is, the probability of selecting first blue then green, plus the probability of selecting first green then blue.

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"The most incomprehensible thing about the world is that it is comprehensible." -Albert Einstein.

mathxyz

Member

From: Brooklyn, NY

Registered: 2024-02-24

Posts: 1,053

Re: Jar of Sweets

For part 2, I would do (9/21)*(12/20) + (12/21)*(9/20). That is, the probability of selecting first blue then green, plus the probability of selecting first green then blue.

Ok. Will do. Thanks again for the set up.