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29 Jan 2016, 02:05
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B
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Difficulty:
95%
(hard)
Question Stats:
51% (03:00) correct
49%
(02:42)
wrong
based on 141
sessions
History
Date
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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
A. 1/33
B. 3/33
C. 5/33
D. 9/33
E. 19/33
31 Jan 2016, 06:06
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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
Answer: B
(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
Answer: B
General Discussion
31 Jan 2016, 22:52
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chetan2u
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
