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30 Jan 2018, 08:20
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From a class of 12 students, two students will be randomly chosen simultaneously. If g is the number of girls in the class, what is the value of g?
(1) The probability that two girls will be chosen together is 1/11.
(2) The probability that one boy and one girl will be chosen is 16/33.
(1) The probability that two girls will be chosen together is 1/11.
(2) The probability that one boy and one girl will be chosen is 16/33.
30 Jan 2018, 09:38
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Bunuel
Let the number of girl be g, then the number of boys =12-g...
(1) The probability that two girls will be chosen together is 1/11.
P of first as girl = \(\frac{g}{12}\)..
P of second as girl = \(\frac{(g-1)}{11}\)..
Combined prob = \(\frac{g}{12} * \frac{(g-1)}{11}=\frac{1}{11}\)...
\(g (g-1)=12=4*3\)...
So g is 4
Sufficient..
2) The probability that one boy and one girl will be chosen is 16/33..
Whatever Ans we get can be of boys or girls..
So insufficient..
\(\frac{(12-g)}{12} *\frac{g}{11} * 2=\frac{16}{33}....... (12-g)g=32...\)
So g can be 8 or 4..
Insuff
A
General Discussion
30 Jan 2018, 21:57
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Assume the number of girls be g
(1) Probability of two girls chosen together is 1/11.
(gC2)/12C2=1/11
or g*(g-1)/(12*11) =1/11
or g*(g-1)=12=4*3
Hence, g is 4
Considering only 1 is Sufficient.
2) Probability of one boy and one girl chosen is 16/33..
gC1*(12-g)C1/12C2) = 16/33
Solving, we get : g*(g-1) = 32
So g can be 8 or 4..
Hence, Insufficient.
A
(1) Probability of two girls chosen together is 1/11.
(gC2)/12C2=1/11
or g*(g-1)/(12*11) =1/11
or g*(g-1)=12=4*3
Hence, g is 4
Considering only 1 is Sufficient.
2) Probability of one boy and one girl chosen is 16/33..
gC1*(12-g)C1/12C2) = 16/33
Solving, we get : g*(g-1) = 32
So g can be 8 or 4..
Hence, Insufficient.
A
