Question

A box contains four 10 nf and eight 100 nf capacitors. (a) what is the probability of obtaining at least two 10 nf capacitors if we randomly draw four capacitors from the box with replacement? (b) what is the probability of drawing at least two 10 nf capacitors if we randomly draw four capacitors from the box without replacement?

Answer 1

there are 4 of the 10-nf and 8 of the 100-nf capacitors which is a total of 12 items.

the probability of drawing a 10-nf is: 4 out of 12 = $\frac{1}{3}$

the probability of drawing a 100-nf is: 8 out of 12 =  $\frac{2}{3}$

(a)  "at least two" means: 2 or 3 or 4

Probability of 2 (10's) and 2 (100's):  $\frac{1}{3}$ x $\frac{1}{3}$ x $\frac{2}{3}$ x $\frac{2}{3}$ = $\frac{4}{81}$

Probability of 3 (10's) and 1 (100's):  $\frac{1}{3}$ x $\frac{1}{3}$ x $\frac{1}{3}$ x $\frac{2}{3}$ = $\frac{2}{81}$

Probability of 4 (10's) and 0 (100's):  $\frac{1}{3}$ x $\frac{1}{3}$ x $\frac{1}{3}$ x $\frac{1}{3}$ = $\frac{1}{81}$

2 or 3 or 4: $\frac{4}{81}$ + $\frac{2}{81}$ + $\frac{1}{81}$ = $\frac{7}{81}$

(b) without replacement

Probability of 2 (10's) and 2 (100's):  $\frac{4}{12}$ x $\frac{3}{11}$ x $\frac{8}{10}$ x $\frac{7}{9}$ = $\frac{4 x 3 x 8 x 7}{12 x 11 x 10 x 9}$

Probability of 3 (10's) and 1 (100's):  $\frac{4}{12}$ x $\frac{3}{11}$ x $\frac{2}{10}$ x $\frac{8}{9}$ = $\frac{4 x 3 x 2 x 8}{12 x 11 x 10 x 9}$

Probability of 4 (10's) and 0 (100's):  $\frac{4}{12}$ x $\frac{3}{11}$ x $\frac{2}{10}$ x $\frac{1}{9}$ = $\frac{4 x 3 x 2 x 1}{12 x 11 x 10 x 9}$

2 or 3 or 4: $\frac{4 x 3 x 8 x 7}{12 x 11 x 10 x 9}$ + $\frac{4 x 3 x 2 x 8}{12 x 11 x 10 x 9}$ + $\frac{4 x 3 x 2 x 1}{12 x 11 x 10 x 9}$ = $\frac{672 + 192 + 24}{12 x 11 x 10 x 9}$ =  $\frac{888}{12 x 11 x 10 x 9}$  =  $\frac{111}{1485}$

Answer 2

Answer:

there are 4 of the 10-nf and 8 of the 100-nf capacitors which is a total of 12 items.

Step-by-step explanation: