Sites: Global Q&A | Wits | MathsGee Club | Joburg Libraries | StartUps | Zimbabwe | OER

MathsGee is Zero-Rated (You do not need data to access) on: Telkom |Dimension Data | Rain | MWEB

1 like 0 dislike
338 views
In a factory there are 100 units of a certain product, 5 of which are defective. We pick three units from the 100 units at random. What is the probability that none of them are defective?
| 338 views

0 like 0 dislike
Let us define $A_{i}$ as the event that the $i$ th chosen unit is not defective, for $i=1,2,3$. We are interested in $P\left(A_{1} \cap A_{2} \cap A_{3}\right)$. Note that
$$P\left(A_{1}\right)=\frac{95}{100}$$
Given that the first chosen item was good, the second item will be chosen from 94 good units and 5 defective units, thus
$$P\left(A_{2} \mid A_{1}\right)=\frac{94}{99}$$
Given that the first and second chosen items were okay, the third item will be chosen from 93 good units and 5 defective units, thus
$$P\left(A_{3} \mid A_{2}, A_{1}\right)=\frac{93}{98}$$
Thus, we have
\begin{aligned} P\left(A_{1} \cap A_{2} \cap A_{3}\right) &=P\left(A_{1}\right) P\left(A_{2} \mid A_{1}\right) P\left(A_{3} \mid A_{2}, A_{1}\right) \\ &=\frac{95}{100} \frac{94}{99} \frac{93}{98} \\ &=0.8560 \end{aligned}

As we will see later on, another way to solve this problem is to use counting arguments
by Bronze Status (9,942 points)

0 like 0 dislike
1 like 0 dislike
1 like 0 dislike
0 like 0 dislike
0 like 0 dislike
2 like 0 dislike
1 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
1 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
1 like 0 dislike
2 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
1 like 0 dislike
0 like 0 dislike
2 like 0 dislike
1 like 0 dislike
2 like 0 dislike
1 like 0 dislike
0 like 0 dislike
2 like 1 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
1 like 0 dislike