Dear Students, based on discussion in theory lecture, you are given an assignment on Probability.
As I have checked the solutions submitted by you, I suggest you to revise them as:
Number of outcomes in S can be calculated as $10 \choose 4$ $=\dfrac{10!}{4! \times 6!}=210$. (These include 0 refurbished, or 1 refurbished, or 2 refurbished, or 3 refurbished computers respectively)
Let A denotes set of all possible outcomes when four computers are selected at random and two of them are refurbished.
Number of outcomes in A can be calculated as $3 \choose 2$ $7 \choose 2$ $=\dfrac{3!}{2! \times 1!}$ $\times \dfrac{7!}{2! \times 5!} = 63$
The required probability is therefore $\dfrac{63}{210}=\dfrac{3}{10}=0.3$
Let B denotes set of all possible outcomes when four computers are selected at random and exactly one of them is refurbished.
Number of outcomes in B can be calculated as $3 \choose 1$ $7 \choose 3$ $=\dfrac{3!}{1! \times 2!}$ $\times \dfrac{7!}{3! \times 4!} = 105$
The required probability is therefore $\dfrac{105}{210}=\dfrac{1}{2}=0.5$
As I have checked the solutions submitted by you, I suggest you to revise them as:
- There are 10 computers in a store. Among them, 7 are brand new and 3 are refurbished. Four computers are purchased for a student lab. From the first look, they are indistinguishable, so the four computers are selected at random. Compute the probability that among the chosen computers,
- two are refurbished
- exactly one is refurbished
Number of outcomes in S can be calculated as $10 \choose 4$ $=\dfrac{10!}{4! \times 6!}=210$. (These include 0 refurbished, or 1 refurbished, or 2 refurbished, or 3 refurbished computers respectively)
Let A denotes set of all possible outcomes when four computers are selected at random and two of them are refurbished.
Number of outcomes in A can be calculated as $3 \choose 2$ $7 \choose 2$ $=\dfrac{3!}{2! \times 1!}$ $\times \dfrac{7!}{2! \times 5!} = 63$
The required probability is therefore $\dfrac{63}{210}=\dfrac{3}{10}=0.3$
Let B denotes set of all possible outcomes when four computers are selected at random and exactly one of them is refurbished.
Number of outcomes in B can be calculated as $3 \choose 1$ $7 \choose 3$ $=\dfrac{3!}{1! \times 2!}$ $\times \dfrac{7!}{3! \times 4!} = 105$
The required probability is therefore $\dfrac{105}{210}=\dfrac{1}{2}=0.5$
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