Question

The probability that a battery will last 10 hr or more is 0.8, and the probability that it will last 15 hr or more is 0.11. Given that a battery has lasted 10 hr, find the probability that it will last 15 hr or more

Answer 1

What do you know about conditional probability and Bayes Theorem?

Answer 2

not much. Does that have anything to do with the problem?

Answer 3

we can do this

Answer 4

put A = the battery lasts 10 hours or more, so \(P(A)=.8\) and B = the batter lasts 15 hours or more so \(P(B)=.11\)

Answer 5

by definition \(P(A|B)=\frac{P(A\cap B )}{P(B)}\) so we only need \(P(A\cap B)\)

Answer 6

but since \(B\subset A\) we know \(P(A\cap B)=P(B)\) so your answer is \(\frac{P(B)}{P(A)}\)

Answer 7

sorry i messed up with the A and B
you want
\[P(B|A)=\frac{P(A\cap B)}{P(A)}\]

Answer 8

to make a long story short your answer is
\[\frac{.11}{.8}\]

Answer 9

What I've been thinking is that
probability = ''desired outcome'' / ''possible outcome''
In this case, desired outcome have a probability of 0.11
possible outcome is the given condition, that is 0.8
Put them back, then I'll get P required = 0.11/0.8.

Actually, I don't understand the third step satellite73 has written. My probability sucks.

Answer 10

satallite73 was right. Thank you!