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Two children, one from each grade, will be chosen at random from 4 chi

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Two children, one from each grade, will be chosen at random from 4 chi  [#permalink]

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New post 26 Apr 2017, 08:39
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Two children, one from each grade, will be chosen at random from 4 children from the 5th grade class and 4 children from the 4th grade class to win a prize. Each class has 3 eligible girls and one eligible boy. What is the probability that one boy and one girl will be chosen?

A. 1/2
B. 3/8
C. 3/5
D. 1/4
E. 1/8

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Re: Two children, one from each grade, will be chosen at random from 4 chi  [#permalink]

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New post 26 Apr 2017, 12:27
Choose 1 from each group. chose a boy from 5ht grade and a girl from 4th grade OR vice versa.

Probability of choosing a boy from 3girls & 1boy set is 1/4
Probability of choosing a girl from 3girls & 1boy set is 3/4

with this we can solve as below:

5th..and 4th

3G1B and 3G1B

1B...and 1G = 1/4 * 3/4
OR
1G...and 1B = 3/4 * 1/4

(1/4 * 3/4) + (3/4 * 1/4)

2 * 1/4 * 3/4

3/8 - B should be correct.

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Re: Two children, one from each grade, will be chosen at random from 4 chi  [#permalink]

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New post 19 May 2017, 09:28
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HKD1710 wrote:
Choose 1 from each group. chose a boy from 5ht grade and a girl from 4th grade OR vice versa.

Probability of choosing a boy from 3girls & 1boy set is 1/4
Probability of choosing a girl from 3girls & 1boy set is 3/4

with this we can solve as below:

5th..and 4th

3G1B and 3G1B

1B...and 1G = 1/4 * 3/4
OR
1G...and 1B = 3/4 * 1/4


(1/4 * 3/4) + (3/4 * 1/4)

2 * 1/4 * 3/4

3/8 - B should be correct.


This is a rational method- though I used the binomial coefficient instead of adding (1/4) and (3/4) twice - 2 (1/4)(3/4) - the binomial coefficient, 2 , simply represents two ways that this order can appear

Btw Bunuel what is the combinatorics method of solving this problem- say if we have 8c2 as the denominator?
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Re: Two children, one from each grade, will be chosen at random from 4 chi  [#permalink]

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New post 19 May 2017, 10:53
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Nunuboy1994 wrote:
HKD1710 wrote:
Choose 1 from each group. chose a boy from 5ht grade and a girl from 4th grade OR vice versa.

Probability of choosing a boy from 3girls & 1boy set is 1/4
Probability of choosing a girl from 3girls & 1boy set is 3/4

with this we can solve as below:

5th..and 4th

3G1B and 3G1B

1B...and 1G = 1/4 * 3/4
OR
1G...and 1B = 3/4 * 1/4


(1/4 * 3/4) + (3/4 * 1/4)

2 * 1/4 * 3/4

3/8 - B should be correct.


This is a rational method- though I used the binomial coefficient instead of adding (1/4) and (3/4) twice - 2 (1/4)(3/4) - the binomial coefficient, 2 , simply represents two ways that this order can appear

Btw Bunuel what is the combinatorics method of solving this problem- say if we have 8c2 as the denominator?


Two children, one from each grade, will be chosen at random from 4 children from the 5th grade class and 4 children from the 4th grade class to win a prize. Each class has 3 eligible girls and one eligible boy. What is the probability that one boy and one girl will be chosen?

A. 1/2
B. 3/8
C. 3/5
D. 1/4
E. 1/8

\(\frac{C^1_3*C^1_1}{C^1_4*C^1_4}+\frac{C^1_1*C^1_3}{C^1_4*C^1_4}=\frac{3}{16}+\frac{3}{16}=\frac{3}{8}\).

Answer: B.

Hope it's clear.
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Re: Two children, one from each grade, will be chosen at random from 4 chi  [#permalink]

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New post 07 Jun 2017, 00:46
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Bunuel
Hi
I used {(3C1)*1}/8C2 thinking that 1 Girl will be selected and 1 boy will be selected as favorable for Grade 4 and 8C2 as total (selecting 2 out of 8).
Just let me know what wrong if I use 8C2.
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Two children, one from each grade, will be chosen at random from 4 chi  [#permalink]

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New post 07 Jun 2017, 02:27
4
As per my understanding,

It is wrong to use 8c2 because you need a single student from each class.
So the denominator of the probability should read 4c1*4c1

Also for numerator, you may have 1 girl from the 3 girls in Grade4 or Grade5
So, it must be 2*3c1*1

Hence the probability is \(\frac{(2*3c1*1)}{(4c1 * 4c1)}\) which gives you \(\frac{3}{8}\)

Hope it helps!
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Re: Two children, one from each grade, will be chosen at random from 4 chi  [#permalink]

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New post 07 Jun 2017, 07:30
1
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Bunuel wrote:
Two children, one from each grade, will be chosen at random from 4 children from the 5th grade class and 4 children from the 4th grade class to win a prize. Each class has 3 eligible girls and one eligible boy. What is the probability that one boy and one girl will be chosen?

A. 1/2
B. 3/8
C. 3/5
D. 1/4
E. 1/8


We can select 1 boy and 1 girl in two ways:
1) select boy from 4th-grade class and girl from 5th grade class
2) select girl from 4th-grade class and boy from 5th grade class

So, P(1 boy and 1 girl) = P(boy from 4th grade AND girl from 5th grade OR girl from 4th grade AND boy from 5th grade)
= P(boy from 4th grade) x P(girl from 5th grade) + P(girl from 4th grade) x P(boy from 5th grade)
= 1/4 x 3/4 + 3/4 x 1/4
= 3/16 + 3/16
= 6/16
= 3/8

Answer:

Cheers,
Brent
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Re: Two children, one from each grade, will be chosen at random from 4 chi  [#permalink]

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New post 10 Jun 2017, 07:17
1
Bunuel wrote:
Two children, one from each grade, will be chosen at random from 4 children from the 5th grade class and 4 children from the 4th grade class to win a prize. Each class has 3 eligible girls and one eligible boy. What is the probability that one boy and one girl will be chosen?

A. 1/2
B. 3/8
C. 3/5
D. 1/4
E. 1/8


There are 2 separate scenarios that will fulfill choosing one boy and one girl.

Scenario 1: a girl from 5th grade and a boy from 4th grade. The probability of this is:

P(girl from 5th and boy from 4th) = 3/4 x 1/4 = 3/16.

Scenario 2: a boy from 5th grade and a girl from 4th grade. The probability of this is:

P(boy from 5th and girl from 4th) = 1/4 x 3/4 = 3/16.

Either of the scenarios satisfies the requirement, so the respective probabilities are added. Thus, the total probability is 3/16 + 3/16 = 6/16 = 3/8.

Answer: B
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Re: Two children, one from each grade, will be chosen at random from 4 chi  [#permalink]

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Re: Two children, one from each grade, will be chosen at random from 4 chi   [#permalink] 29 Oct 2018, 01:37
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