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Difficulty: Very hardIndependent and Dependent Events

A quality control inspector tests items from a batch containing nn high-grade items and 44 low-grade items, where n>1n > 1. Two items are selected sequentially. If the first item selected is high-grade, it is kept out of the batch for testing, and additionally 11 low-grade item is also removed from the batch before the second selection. If the probability of selecting two high-grade items in succession is 38\frac{3}{8}, what is the value of nn?

Answer: 6

Answer

The initial number of high-grade items nn is 66.
Because the outcome of the first draw alters both the count of high-grade items and the count of low-grade items, the two draws are dependent events. The probability of drawing a high-grade item on the first draw is nn+4\frac{n}{n+4}. Following this successful outcome, the remaining pool consists of n−1n-1 high-grade items and 33 low-grade items, yielding a total of n+2n+2 items. The probability of a second high-grade selection is therefore n−1n+2\frac{n-1}{n+2}. Multiplying these probabilities gives n(n−1)(n+4)(n+2)=38\frac{n(n-1)}{(n+4)(n+2)} = \frac{3}{8}. Solving the resulting quadratic equation 5n2−26n−24=05n^2 - 26n - 24 = 0 gives n=6n = 6 (since nn must be a positive integer).

Step-by-Step Solution

1
Calculate the probability of selecting a high-grade item on the first draw.
P(H1)=nn+4P(H_1) = \frac{n}{n+4}
The initial batch contains nn high-grade items out of a total of n+4n+4 items.
2
Determine the conditional sample space for the second draw after event H1H_1 occurs.
High-grade items remaining = n−1n-1; low-grade items remaining = 4−1=34-1=3; total remaining items = (n−1)+3=n+2(n-1)+3 = n+2.
Drawing a high-grade item removes 11 high-grade item and 11 additional low-grade item from the batch.
3
Calculate the conditional probability of selecting a high-grade item on the second draw.
P(H2∣H1)=n−1n+2P(H_2 | H_1) = \frac{n-1}{n+2}
There are now n−1n-1 high-grade items out of n+2n+2 total remaining items.
4
Set up and solve the algebraic probability equation.
n(n−1)(n+4)(n+2)=38  ⟹  8n2−8n=3(n2+6n+8)  ⟹  5n2−26n−24=0  ⟹  (5n+4)(n−6)=0  ⟹  n=6\frac{n(n-1)}{(n+4)(n+2)} = \frac{3}{8} \implies 8n^2 - 8n = 3(n^2 + 6n + 8) \implies 5n^2 - 26n - 24 = 0 \implies (5n+4)(n-6) = 0 \implies n = 6.
Because n>1n > 1 is required, the negative fraction root −45-\frac{4}{5} is discarded.

Key Concept

Probability of dependent events with non-standard sample space modifications
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Topics

Independent and Dependent Events

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