Two bolts, two nuts and three needles are in a box. Two parts are choosen at random from the box. What is…
Two bolts, two nuts and three needles are in a box. Two parts are choosen at random from the box. What is the probability that one is a
bolt and one is a needle?
$\frac{2}{21}$
$\frac{4}{21}$
$\frac{6}{21}$
$\frac{12}{21}$
Solution
2 Bolts, 2 Nuts, 3 Needles Total number of possible outcomes, when we choose 2 parts
$n(S)={ }^7 C_2=\frac{7 \times 6}{1 \times 2}=21$
Let E be the event that the parts drawn are bolt and needle.
$\begin{array}{ll}\therefore & n(E)={ }^2 C_1 \times{ }^3 C_1=6 \\ \therefore & P(E)=\frac{n(E)}{n(S)}=\frac{6}{21}\end{array}$