There are 2 green, 3 red, and 2 blue balls in a box. 4 are drawn at random without replacement. What is the probability that of the 4 drawn balls two are red, 1 is green, and 1 is blue?
I solved it this way:
3/7 * 2/6 * 2/5 * 2/4 = 1/35
thinking that the solution is: p(r) * p(r) * p(g) * p(b)
and the numbers decrease after each selection because the balls are picked without replacement.
The answer however, is 12/35. Can someone please explain why?
Difficult probability
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- joannabanana
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Solved it using combinations:
Looks like this: [(3C2)(2C1)(2C1)] / (7C4) = 12/35
Explanation of Terms:
1st term- # ways to arrange 3 Red Marbles in 2 picks: 3C2 = 3!/(1!)(2!) = 3
2nd term - # ways to arrange 2 Blue marbles in 1 pick: 2C1 = 2!/(1!)(1!) = 2
3rd term- # ways to arrange 2 Green marbles in 1 pick: 2C1 = 2!/(1!)(1!) = 2
Denominator- How many ways to arrange 7 marbles in 4 picks: 7C4= 7!/(4!)(3!) = 35
So 35 ways to arrange 4 marbles out of 7.
(3)(2)(2)/35 = 12/35
Your method assumed the marbles were picked in the order of RED, RED, GREEN, BLUE , but there are actually 12 ways we could pick these:
RRGB RGBR RBGR RRBG RBRG RGRB
GRRB GBRR GRBR BGRR BRGR BRRG
Looks like this: [(3C2)(2C1)(2C1)] / (7C4) = 12/35
Explanation of Terms:
1st term- # ways to arrange 3 Red Marbles in 2 picks: 3C2 = 3!/(1!)(2!) = 3
2nd term - # ways to arrange 2 Blue marbles in 1 pick: 2C1 = 2!/(1!)(1!) = 2
3rd term- # ways to arrange 2 Green marbles in 1 pick: 2C1 = 2!/(1!)(1!) = 2
Denominator- How many ways to arrange 7 marbles in 4 picks: 7C4= 7!/(4!)(3!) = 35
So 35 ways to arrange 4 marbles out of 7.
(3)(2)(2)/35 = 12/35
Your method assumed the marbles were picked in the order of RED, RED, GREEN, BLUE , but there are actually 12 ways we could pick these:
RRGB RGBR RBGR RRBG RBRG RGRB
GRRB GBRR GRBR BGRR BRGR BRRG
- GMATGuruNY
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You determined P(RRBG) = 1/35.joannabanana wrote:There are 2 green, 3 red, and 2 blue balls in a box. 4 are drawn at random without replacement. What is the probability that of the 4 drawn balls two are red, 1 is green, and 1 is blue?
I solved it this way:
3/7 * 2/6 * 2/5 * 2/4 = 1/35
thinking that the solution is: p(r) * p(r) * p(g) * p(b)
and the numbers decrease after each selection because the balls are picked without replacement.
The answer however, is 12/35. Can someone please explain why?
But RRBG is only 1 way to get a good outcome. To account for all the ways we could get 2 R's, 1 B and 1 G, we need to multiply by the number of ways to arrange RRBG = 4!/2! = 12.
12*1/35 = 12/35.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
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