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Probability Question?
So my teacher just made a very unique if not brand new question that's never been asked by my other teachers before.
A man is given a chance to try out 12 TVs, 8 of which are not broken. A test is conducted to sort the TVs. What is the probability that on the 6th try the TV tested is :
A. the third that is broken
B. the fourth that is broken.
Just crosschecking with my teacher's answer and if possible give justification for answer.
3 Answers
- M3Lv 710 years agoFavourite answer
this is drawal w/o replacement,
8 ok, 4 bad, 12 total,
6 TVS tested
qA.
2 bad & 3 ok must be there in 5 tries followed by a bad on 6th try
Pr = (4c2*8c3 / 12c5 ) * 2/7 = 4/33 or 12.12% <--------
qB
3 bad & 2 ok must be there in 5 tested followed by a bad on 6th try
Pr = (4c3*8c2)/12c5) *1/7 = 2/99 or 2.02% <--------
- cidyahLv 710 years ago
A)
P(broken TV) = 4/12=1/3 = .3333
On the first 5 tries, 2 are broken and on the 6th try 1 TV is broken.
P(2 broken TV in 5) = 5C2 (1/3)^2 (4/3)^3 = (0.329202) ---- binomial distribution
P( 2 broken TV in 5 and the 3rd broken TV in the 6th) = (0.329202)(.3333) = 0.1097
B)
On the first 5 tries, 3 are broken and on the 6th try 1 TV is broken.
P(3 broken TV in 5) = 5C3 (1/3)^3 (4/3)^2 = (0.329202) ---- binomial distribution
P( 3 broken TV in 5 and the 4th broken TV in the 6th) = (0.164576)(.3333) = 0.05485
