Question

A display case contains thirty-five gems, of which ten are real diamonds and twenty-five are fake diamonds. A burglar removes four gems at random, one at a time and without replacement. What is the probability that the last gem she steals is the second real diamond in the set of four?

Answer 1

Answer:0.0128

Step-by-step explanation:

Given

There are 35 gems out of which there are 10 real diamonds and 25 are fake diamonds.

Probability that  the last gem burglar steals is second real diamond in the set of four

there can three possibilities

a)Real , fake, fake , real

b)fake , real , fake , real

c)fake, fake, real ,real

a)$P=\frac{10}{35}\times \frac{25}{34}\times \frac{24}{33}\times \frac{9}{32}=\frac{25\times 24\times 10\times 9}{35\times 34\times 33\times 32}$                    

b)$P=\frac{25}{35}\times \frac{10}{34}\times \frac{24}{33}\times \frac{9}{32}=\frac{25\times 24\times 10\times 9}{35\times 34\times 33\times 32}$

c)$P=\frac{25}{35}\times \frac{24}{34}\times \frac{10}{33}\times \frac{9}{32}=\frac{25\times 24\times 10\times 9}{35\times 34\times 33\times 32}$

$P_{total}=3\times =\frac{25\times 24\times 10\times 9}{35\times 34\times 33\times 32}=\frac{54000}{1256640}=0.0128$