How to solve this Sec 4 probability Question part (c) & (d). Appreciate i anybody can help
Sec 4 probability Question
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student101  YellowBelt
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Re: Sec 4 probability Question
by Lavina » Fri Oct 02, 2015 6:33 pm
C) 3(3/5(2/5)^4 + 3/5(2/5)^5)
D) (2/5)^(n1) x 3/5
D) (2/5)^(n1) x 3/5

Lavina  GreenBelt
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Re: Sec 4 probability Question
by student101 » Sat Oct 03, 2015 3:16 pm
Thanks Lavina. I understood the answer for d. But still confused with answer for c. Can give a hint for c?

student101  YellowBelt
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Re: Sec 4 probability Question
by XXXX » Sat Oct 03, 2015 4:37 pm
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.
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.

XXXX  BlueBelt
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Re: Sec 4 probability Question
by student101 » Mon Oct 05, 2015 9:28 am
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
Now I got both of your concept. Thanks a lot to xxxx and Lavina

student101  YellowBelt
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Pls help with this O level matrix question
by student101 » Mon Oct 12, 2015 10:33 pm
I dont understand part b & c of this O level matrix Q. Pls do help me. Thanks

student101  YellowBelt
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