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jackfr2
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An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles. If four 04 Nov 2019, 07:17
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Question Stats:

71% (02:29) correct 29% (02:45) wrong based on 236 sessions

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An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles. If four marbles are picked up at random, what is the probability that 1 is green, 2 are blue and 1 is red?
a)13/35
(b) 24/455
(c) 11/15
(d) 1/13
(e) 2/455
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B
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chetan2u
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An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles. If four 04 Nov 2019, 07:56
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jackfr2
An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles. If four marbles are picked up at random, what is the probability that 1 is green, 2 are blue and 1 is red?
a)13/35
(b) 24/455
(c) 11/15
(d) 1/13
(e) 2/455


Total marbles = 6+4+2+3=15..Ways to select 4 out of them = 15C4=\(\frac{15*14*13*12}{4*3*2}=15*7*13\)

Ways to select 1 green, 2 blue and 1 red = 2C1*4C2*6C1=2*6*6

Probability = \(\frac{2*6*6}{15*7*13}=\frac{24}{455}\)

B
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jackfr2
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An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles. If four 04 Nov 2019, 09:30
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chetan2u
jackfr2
An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles. If four marbles are picked up at random, what is the probability that 1 is green, 2 are blue and 1 is red?
a)13/35
(b) 24/455
(c) 11/15
(d) 1/13
(e) 2/455


Total marbles = 6+4+2+3=15..Ways to select 4 out of them = 15C4=\(\frac{15*14*13*12}{4*3*2}=15*7*13\)

Ways to select 1 green, 2 blue and 1 red = 2C1*4C2*6C1=2*6*6

Probability = \(\frac{2*6*6}{15*7*13}=\frac{24}{455}\)

B

probability:2*4*3*6/15*14*13*12=2/455

Why can't I solve this math using probability approach??
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chetan2u
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An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles. If four 04 Nov 2019, 17:00
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jackfr2
chetan2u
jackfr2
An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles. If four marbles are picked up at random, what is the probability that 1 is green, 2 are blue and 1 is red?
a)13/35
(b) 24/455
(c) 11/15
(d) 1/13
(e) 2/455


Total marbles = 6+4+2+3=15..Ways to select 4 out of them = 15C4=\(\frac{15*14*13*12}{4*3*2}=15*7*13\)

Ways to select 1 green, 2 blue and 1 red = 2C1*4C2*6C1=2*6*6

Probability = \(\frac{2*6*6}{15*7*13}=\frac{24}{455}\)

B

probability:2*4*3*6/15*14*13*12=2/455

Why can't I solve this math using probability approach??

You can solve it via probability method too.
Here 1G, 2B and 1R can be picked up in 4!/2! Or 12 ways.
So multiply your result by 12...12*2/455=24/455
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niekhiill99
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An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles. If four 14 Dec 2024, 13:54
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how is 2*6*6=24?
It ahould be 72 i guess! Please help
chetan2u
jackfr2
An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles. If four marbles are picked up at random, what is the probability that 1 is green, 2 are blue and 1 is red?
a)13/35
(b) 24/455
(c) 11/15
(d) 1/13
(e) 2/455


Total marbles = 6+4+2+3=15..Ways to select 4 out of them = 15C4=\(\frac{15*14*13*12}{4*3*2}=15*7*13\)

Ways to select 1 green, 2 blue and 1 red = 2C1*4C2*6C1=2*6*6

Probability = \(\frac{2*6*6}{15*7*13}=\frac{24}{455}\)

B
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niekhiill99
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An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles. If four 14 Dec 2024, 13:54
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how is 2*6*6=24?
It ahould be 72 i guess! Please help
chetan2u
jackfr2
An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles. If four marbles are picked up at random, what is the probability that 1 is green, 2 are blue and 1 is red?
a)13/35
(b) 24/455
(c) 11/15
(d) 1/13
(e) 2/455


Total marbles = 6+4+2+3=15..Ways to select 4 out of them = 15C4=\(\frac{15*14*13*12}{4*3*2}=15*7*13\)

Ways to select 1 green, 2 blue and 1 red = 2C1*4C2*6C1=2*6*6

Probability = \(\frac{2*6*6}{15*7*13}=\frac{24}{455}\)

B
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Shayan08
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An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles. If four 17 Dec 2024, 06:42
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One of the 6s in the numerator and the 15 in denominator were divided by 3. This gives the numerator 2*2*6=24
niekhiill99
how is 2*6*6=24?
It ahould be 72 i guess! Please help
chetan2u
jackfr2
An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles. If four marbles are picked up at random, what is the probability that 1 is green, 2 are blue and 1 is red?
a)13/35
(b) 24/455
(c) 11/15
(d) 1/13
(e) 2/455


Total marbles = 6+4+2+3=15..Ways to select 4 out of them = 15C4=\(\frac{15*14*13*12}{4*3*2}=15*7*13\)

Ways to select 1 green, 2 blue and 1 red = 2C1*4C2*6C1=2*6*6

Probability = \(\frac{2*6*6}{15*7*13}=\frac{24}{455}\)

B
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