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# The probability of picking 2 red balls one after another without repla

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Math Expert
Joined: 02 Aug 2009
Posts: 8326
The probability of picking 2 red balls one after another without repla  [#permalink]

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29 Jan 2016, 02:05
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Difficulty:

75% (hard)

Question Stats:

55% (03:12) correct 45% (02:39) wrong based on 92 sessions

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The probability of picking 2 red balls one after another without replacement from a bag containing only red and white balls is 14/33. What will be the probability of picking 2 white balls one after another w/o replacement from that bag, given that it can hold at the most 20 balls?
A. 1/33
B. 3/33
C. 5/33
D. 9/33
E. 19/33

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Marshall & McDonough Moderator
Joined: 13 Apr 2015
Posts: 1683
Location: India
Re: The probability of picking 2 red balls one after another without repla  [#permalink]

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31 Jan 2016, 06:06
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4
Probability of picking 2 red balls = 14/33
(R/R+W)*(R-1)/(R+W-1) = 14/33 = (2*7)/(3*11) = (2/3)*(7/11) --> Once the value is reduced to this level, it should be easy to find out the number of red balls.

1) Considering the above expression, both numerator and denominator must be consecutive.
2) Lets multiply both numerator and denominator with the same integer so that the value remains unaffected.
3) We can't multiply 7/11 as we need a consecutive term in both numerator and denominator. So lets multiply (2/3) to make the numerator and denominator consecutive.

(2/3)*(2/2) = 4/6 --> 4*7/6*11 --> Numerator and denominator not consecutive
(2/3)*(3/3) = 6/9 --> 6*7/9*11 --> Only numerator is consecutive
(2/3)*(4/4) = 8/12 --> 8*7/12*11 --> Numerator and denominator are consecutive here

Number of red balls, R = 8.
Total number of balls, R + W = 12 (This value is < 20, satisfying the condition)
Number of white balls, W = 12 - 8 = 4

Probability of picking 2 white balls w/o replacement = (W/R+W)*(W-1)/(R+W-1) = (4/12)*(3/11) = 3/33

##### General Discussion
Math Expert
Joined: 02 Aug 2009
Posts: 8326
Re: The probability of picking 2 red balls one after another without repla  [#permalink]

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31 Jan 2016, 22:52
3
chetan2u wrote:
The probability of picking 2 red balls one after another without replacement from a bag containing only red and white balls is 14/33. What will be the probability of picking 2 white balls one after another w/o replacement from that bag, given that it can hold at the most 20 balls?
A. 1/33
B. 3/33
C. 5/33
D. 9/33
E. 19/33

Hi,

Lets see what info is being provided by the Q...
1) The bag contains only red and white balls.
2) Prob of picking two red balls w/o replacement =14/33
3) A critical info it can hold at the most 20 balls...

what does " Prob of picking two red balls w/o replacement =14/33" mean if there are R red balls and W white balls..
$$\frac{R}{{R+W}} *\frac{{R-1}}{{R+W-1}}=\frac{14}{33}$$..
or $$\frac{R{(R-1)}}{{(R+W)}{(R+W-1)}}= \frac{14}{33}$$.....

WHAT DOES RHS MEAN
14/33.. this means the actual numbers can be taken as 14x/33x, where x is the common term

WHAT DOES LHS MEAN
$$\frac{R{(R-1)}}{{(R+W)}{(R+W-1)}}$$..
it means both nemerator and denominator are multiple of consecutive numbers..

from above two points we can say ..
14x is the product of two consecutive integers..
2*7*x..
33x is also the product of two consecutive integers..
3*11*x

as we see, x can be fitted as 4 to get numerator as 7*8 and denominator as 11*12...
so $$\frac{R{(R-1)}}{{(R+W)}{(R+W-1)}}$$..= $$\frac{{(8*7)}}{{(12*11)}}$$..
.... or $$\frac{8{(8-1)}}{{(8+4)}{(8+4-1)}}$$..
thus R = 8 and w=4..

thus prob of picking 2 white balls one after another w/o replacement from that bag=4/12*3/11=1/11=3/33..
ans B
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Joined: 22 Oct 2015
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The probability of picking 2 red balls one after another without repla  [#permalink]

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10 Feb 2016, 16:16
chetan2u wrote:
chetan2u wrote:
The probability of picking 2 red balls one after another without replacement from a bag containing only red and white balls is 14/33. What will be the probability of picking 2 white balls one after another w/o replacement from that bag, given that it can hold at the most 20 balls?
A. 1/33
B. 3/33
C. 5/33
D. 9/33
E. 19/33

Hi,

Lets see what info is being provided by the Q...
1) The bag contains only red and white balls.
2) Prob of picking two red balls w/o replacement =14/33
3) A critical info it can hold at the most 20 balls...

what does " Prob of picking two red balls w/o replacement =14/33" mean if there are R red balls and W white balls..
$$\frac{R}{{R+W}} *\frac{{R-1}}{{R+W-1}}=\frac{14}{33}$$..
or $$\frac{R{(R-1)}}{{(R+W)}{(R+W-1)}}= \frac{14}{33}$$.....

WHAT DOES RHS MEAN
14/33.. this means the actual numbers can be taken as 14x/33x, where x is the common term

WHAT DOES LHS MEAN
$$\frac{R{(R-1)}}{{(R+W)}{(R+W-1)}}$$..
it means both nemerator and denominator are multiple of consecutive numbers..

from above two points we can say ..
14x is the product of two consecutive integers..
2*7*x..
33x is also the product of two consecutive integers..
3*11*x

as we see, x can be fitted as 4 to get numerator as 7*8 and denominator as 11*12...
so $$\frac{R{(R-1)}}{{(R+W)}{(R+W-1)}}$$..= $$\frac{{(8*7)}}{{(12*11)}}$$..
.... or $$\frac{8{(8-1)}}{{(8+4)}{(8+4-1)}}$$..
thus R = 8 and w=4..

thus prob of picking 2 white balls one after another w/o replacement from that bag=4/12*3/11=1/11=3/33..
ans B

3) A critical info it can hold at the most 20 balls...

You mention this as a critical info yet I fail to see how exactly this critical info was used in the process of determining the answer. Could you please elaborate?
Marshall & McDonough Moderator
Joined: 13 Apr 2015
Posts: 1683
Location: India
The probability of picking 2 red balls one after another without repla  [#permalink]

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10 Feb 2016, 18:04
1
Hi ZaydenBond,

According to the question, the total number of balls should be less than or equal to 20.
We have obtained the total number of balls as 12 which is the only number satisfying the given critical info.
If the number 20 was not given, we could have other possible solutions as the given ratio is 14x/33x, where x can take any value.
Manager
Joined: 04 Jun 2015
Posts: 77
Re: The probability of picking 2 red balls one after another without repla  [#permalink]

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10 Jul 2017, 08:28
chetan2u wrote:
The probability of picking 2 red balls one after another without replacement from a bag containing only red and white balls is 14/33. What will be the probability of picking 2 white balls one after another w/o replacement from that bag, given that it can hold at the most 20 balls?
A. 1/33
B. 3/33
C. 5/33
D. 9/33
E. 19/33

In the actual test I might just guess this one. Here are my 2 cents though -

Take a look at the wording of last part of the sentence "....it can hold at most 20 balls"
1. Max number of balls that the bag can hold is 20.
2. Two balls are picked w/o replacement.

Now if we take total as 12 see what happens - we are given that R/12 * (R-1)/11 = 14/33
Easy to calculate R = 8, W = 4.

So probability of picking 2 balls white = 4/12 * 3/11 = 1/11 = 3/33
Hence B
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Joined: 24 Nov 2016
Posts: 972
Location: United States
Re: The probability of picking 2 red balls one after another without repla  [#permalink]

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24 Nov 2019, 07:11
chetan2u wrote:
The probability of picking 2 red balls one after another without replacement from a bag containing only red and white balls is 14/33. What will be the probability of picking 2 white balls one after another w/o replacement from that bag, given that it can hold at the most 20 balls?
A. 1/33
B. 3/33
C. 5/33
D. 9/33
E. 19/33

0<t=r+w≤20 (integers)

$$r(r-1)/t(t-1)=14/33=7*2/3*11…r^2-r=14[t(t-1)]/33$$

14 ≠ div by 3 and 11, so t(t-1) must be div 3 and 11
if t=11 and t-1=10 are div by 11 but not by 3;
if t=12 and t-1=11 are div by 11 and 3;

$$r^2-r=14[12(11)]/33…r^2-r-56=0…(r-8)(r+7)=0…r=8, t=12, w=4$$

$$w(w-1)/t(t-1)=4(3)/12(11)=1/11=3/33$$

Ans (B)
Re: The probability of picking 2 red balls one after another without repla   [#permalink] 24 Nov 2019, 07:11
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