Defective bulbs: n
Non-defective bulbs: 10-n
and 0<=n<5.
Two bulbs can be drawn from a lot of 10 in
C^{10}_{2} = \frac{10!}{2!8!}=\frac{(10*9)}{2}=45ways.
(1) Both defective:
Probability(Pick both defective bulbs) = Number of favorable choices/Number of total choices
Number of favorable choices : If we draw both defective bulbs
i.e. {draw 2 balls out of n defective balls} * {0 balls out of (10-n) balls}
The number of ways that can be done is:
C^{n}_{2} * C^{(10-n)}_{0}=> \frac{n!}{(n-2)!2!} * 1\frac{n(n-1)(n-2)!}{(n-2)!2!}=\frac{n(n-1)}{2}Number of favorable choices:
\frac{n(n-1)}{2}Number of total choices: 45
Probability(Pick both defective bulbs) =
\frac{n(n-1)}{(2*45)}Given, Probability(Pick both defective bulbs) = 1/15
\frac{n(n-1)}{90} = \frac{1}{15}n(n-1) = 6n^2-n-6=0Solving the quadratic equation;
n=3, n=-2
Since,
0<=n<5
n = 3
Sufficient.
(2) One defective and one non-defective:
Probability(One defective and one non-defective) = Number of favorable choices/Number of total choices
Number of favorable choices : If we draw exactly one defective and one non-defective bulb
i.e. {draw 1 balls out of n defective balls} * {1 balls out of (10-n) balls}
The number of ways that can be done is:
C^{n}_{1} * C^{(10-n)}_{1}=> \frac{n!}{(n-1)!1!} * \frac{(10-n)!}{(10-n-1)!1!}=> \frac{n(n-1)!}{(n-1)!1!} * \frac{(10-n)(10-n-1)!}{(10-n-1)!1!}n(10-n)Number of favorable choices:
n(10-n)Number of total choices: 45
Probability(One defective and one non-defective) =
\frac{n(10-n)}{45}Given, Probability(One defective and one non-defective) = 7/15
\frac{n(10-n)}{45} = \frac{7}{15}n(10-n) = 21n^2-10n+21=0Solving the quadratic equation gives:
n=3 and n=7
Since,
0<=n<5
n = 3
Sufficient.
Ans: "D"
_________________
~fluke
Find out what's new at GMAT Club - latest features and updates
close
73% (02:27) correct
