Question

A bag contains six real diamonds and five fake diamonds. If six diamonds are picked from the bag at random, what is the probability that at most four of them are real?

Answer 1

Answer:

0.933 ( approx )

Step-by-step explanation:

Given,

Real diamonds = 6,

Fake diamonds = 5,

Total diamonds = 6 + 5 = 11,

If six diamonds are picked from the bag at random,

So, the total ways = $^{11}C_6$

$=\frac{11!}{6!5!}$

= 462

Also, the number of ways of at most four real diamonds in these 6 diamonds

$=^{6}C_1\times ^{5}C_5 + ^{6}C_2\times ^{5}C_4+^{6}C_3\times ^{5}C_3+^{6}C_4\times ^{5}C_2$

$=6+75+200+150$

$=431$

Hence, the probability that at most four of them are real = $\frac{431}{462}$

$\approx 0.933$

Answer 2

The probability that the diamond picked will be real at the start is:
P1 = 6/11
If the diamond is not replaced back into the container, then the probability that at most four of them are real is:
P = (6/11)(5/10)(4/9)(3/8)(5/7)(6/6)
P = 5/154 or
P = 0.0325 or
P = 3.2468%