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A bag has 5 red balls and 7 white balls. Three balls are drawn randoml
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12 Mar 2023, 10:05
12 Mar 2023, 10:05
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A bag has 5 red balls and 7 white balls. Three balls are drawn randomly from the bag (without replacement). What is the probability that two of the balls drawn are white and one is red?
A. \(\frac{7}{44}\)
B. \(\frac{7}{22}\)
C. \(\frac{21}{44}\)
D. \(\frac{21}{22}\)
E. \(\frac{35}{110}\)
A. \(\frac{7}{44}\)
B. \(\frac{7}{22}\)
C. \(\frac{21}{44}\)
D. \(\frac{21}{22}\)
E. \(\frac{35}{110}\)
Quant Chat Moderator
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Re: A bag has 5 red balls and 7 white balls. Three balls are drawn randoml
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12 Mar 2023, 12:53
12 Mar 2023, 12:53
TBT wrote:
A bag has 5 red balls and 7 white balls. Three balls are drawn randomly from the bag (without replacement). What is the probability that two of the balls drawn are white and one is red?
a. \(\frac{7}{44}\)
b. \(\frac{7}{22}\)
c. \(\frac{21}{44}\)
d. \(\frac{21}{22}\)
e. \(\frac{35}{110}\)
a. \(\frac{7}{44}\)
b. \(\frac{7}{22}\)
c. \(\frac{21}{44}\)
d. \(\frac{21}{22}\)
e. \(\frac{35}{110}\)
Let's assume the balls are drawn in this order
R R W
The probability of drawing a red ball in the first place = \(\frac{7}{12}\)
The probability of drawing a red ball in the second place = \(\frac{6}{11}\)
The probability of drawing a white ball in the third place = \(\frac{5 }{10}\)
In this case, we have taken the order as R R W, however, the order need not be just this. For example, W R R, and R W R also hold true. These are nothing but arrangements of R R W.
Hence, the net probability = \(\frac{7}{12}\) * \(\frac{6}{11}\) * \(\frac{5 }{10}\) * \(\frac{3!}{2!}\)
Net Probability = \(\frac{21}{44}\)
Option C
Re: A bag has 5 red balls and 7 white balls. Three balls are drawn randoml
[#permalink]
12 Mar 2023, 23:07
12 Mar 2023, 23:07
gmatophobia wrote:
TBT wrote:
A bag has 5 red balls and 7 white balls. Three balls are drawn randomly from the bag (without replacement). What is the probability that two of the balls drawn are white and one is red?
a. \(\frac{7}{44}\)
b. \(\frac{7}{22}\)
c. \(\frac{21}{44}\)
d. \(\frac{21}{22}\)
e. \(\frac{35}{110}\)
a. \(\frac{7}{44}\)
b. \(\frac{7}{22}\)
c. \(\frac{21}{44}\)
d. \(\frac{21}{22}\)
e. \(\frac{35}{110}\)
Let's assume the balls are drawn in this order
R R W
The probability of drawing a red ball in the first place = \(\frac{7}{12}\)
The probability of drawing a red ball in the second place = \(\frac{6}{11}\)
The probability of drawing a white ball in the third place = \(\frac{5 }{10}\)
In this case, we have taken the order as R R W, however, the order need not be just this. For example, W R R, and R W R also hold true. These are nothing but arrangements of R R W.
Hence, the net probability = \(\frac{7}{12}\) * \(\frac{6}{11}\) * \(\frac{5 }{10}\) * \(\frac{3!}{2!}\)
Net Probability = \(\frac{21}{44}\)
Option C
Not sure where I’m going wrong
(7C2 x 5C1) / 12 C3
=21/440
Posted from my mobile device
