Question

A firm buys 3 shipments of parts each month. The purchasing agent selects at random from among four in-state suppliers and six out-of-state suppliers. What is the probability that orders are placed with
a) The in-state suppliers only
b) The out-of-state suppliers only
c) At least one in- state suppliers

Answer 1

in-state suppliers (I) $=4$

Out of state suppliers (O) $=6$

Suppliers picked (r) $=3$

Total suppliers (N) $=10$

Probability of selecting 3 suppliers from the total

$$=\frac{_I C_x * _OC_y}{_N C_r}$$

Where 'I' are the in-state suppliers selected and 'O' the number of out of state suppliers

solution a) in-state suppliers only

$$=\frac{4 C_3}{10C_3} = \frac{4}{120}$$

answer: 0.0333

solution b) out of state suppliers only

$$=\frac{6C_3}{10C_3} = \frac{20}{120}$$

answer: 0.1667

solution c) atleast 1 in-state supplier

when in-state supplier $=1$

$$Probability =\frac{4C_1 * 6C_2}{10C_3} = \frac{4*15}{120} = 0.5$$

when in-state suppliers $=2$

$$Probability = \frac{4C_2 * 6C_1}{10C_3} = \frac{6*6}{120}=0.3$$

when in-state suppliers$=3$

$Probability = 0.0333$

Probability of atleast 1 in-state supplier

$$=0.5+0.3+0.0333$$

answer: 0.8333