Sec 4 probability Question

[Lavina]

C) 3(3/5(2/5)^4 + 3/5(2/5)^5)


D) (2/5)^(n-1) x 3/5

[student101]

Thanks Lavina. I understood the answer for d. But still confused with answer for c. Can give a hint for c?

[XXXX]

For each candidate, P(pass) is 21/25 and P(fail) is 4/25. P(pass) = 3/5 + 2/5 x 3/5


Then for 3 candidates taking test, there are 8 possible outcomes (in terms of pass/fail), of which 3 are 1 candidate only passing; 100, 010, 001.

The probability of this outcome can be seen simply in a probability tree and is:
3 x (21/25 x 4/25 x 4/25) = 0.064512.

We multiply because it is a contingent tree, you only follow the paths in the space that give 1 pass only.

Apologies if I am wrong, a bit HAZY today.

[student101]

Thanks xxxx for your detailed solution. I think the answer is right and is same as Lavina’s answer key.

Now I got both of your concept. Thanks a lot to xxxx and Lavina

[student101]

I dont understand part b & c of this O level matrix Q. Pls do help me. Thanks

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