How Do You Calculate the Probability of Bus Frame Cracks?

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This discussion focuses on calculating the probability of bus frame cracks in a specific scenario involving 25 buses, of which 8 have developed cracks. The combinatorial calculations reveal that there are 53,130 ways to select a sample of 5 buses. Furthermore, the number of samples containing exactly 4 buses with visible cracks is determined to be 1,190. The probability of randomly selecting a sample of 5 buses with exactly 4 having cracks is calculated as 0.022, while the probability of selecting at least 4 cracked buses is derived from the combination of two scenarios.

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Shortly after being put into service, some buses manufactured by a certain company have developed cracks on the underside of the main frame. Suppose a particular city has 25 of these buses, and cracks have actually appeared in 8 of them.

(i) How many ways are there to select a sample of 5 buses from the 25 for a thorough inspection?
(ii) How many of these samples of 5 buses contain exactly 4 with visible cracks?
(iii) If a sample of 5 buses is chosen at random, what is the probability that exactly 4 of the 5 will have visible cracks (to 3 dp)?
(iv) If buses are selected as in part (iii), what is the probability that at least 4 of those selected will have visible cracks (to 3 dp)?answers:
(i) 25 buses, 5 choices so: 25C5 = 53130
(ii) 5 buses, 8 broken so: (8C4)(17C1) = 1190
(iii) 53130/1190
(iv)
at least 4, so:

{(8C4)(17C1) + (8C5)(17C0)} / 25C5can you check my answers? thanks.
 
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Apart from putting the answer to (iii) upside down, your answers are correct.

You might get a better response in the "homework" section.
 

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