Question

A box contains 16 ​transistors, 5 of which are defective. if 5 are selected at​ random, find the probability that

Answer 1

Part (a): All are defective
Only one way of selecting the 5 defective transistors:

Number of ways of selections available = 6C5 = 16!/[5!*(15-5)!] = 4368

Probability they are all defective = Number of ways of selecting 5 defectives/Total number of ways possible = 1/4368 ≈ 0.000229

Part (b): None are defective
Total number of non defectives = 16 -5 = 11
Number of ways of selecting 5 non defective = 11C5 = 462 ways
Total number of ways possible = 16C5 = 4368
Probability of selecting 5 non defectives = 462/4368 = 11/104 ≈ 0.1058

Answer 2

A - all defective

B - none defective

$\displaystyle |\Omega|=\binom{16}{5}=\dfrac{16!}{5!11!}=\dfrac{12\cdot13\cdot14\cdot15\cdot16}{2\cdot3\cdot4\cdot5}=4368\\ |A|=1\\ |B|=\binom{11}{5}=\dfrac{11!}{5!6!}=\dfrac{7\cdot8\cdot9\cdot10\cdot11}{2\cdot3\cdot4\cdot5}=462\\\\ P(A)=\dfrac{1}{4368}\approx0.02\%\\ P(B)=\dfrac{462}{480480}=\dfrac{1}{1040}\approx0.1\%$