John plays a game for 60 minutes. For the first 20 mins he has a 1/2 chance of winning and for the last 40 mins he has a 1/5 chance of winning. What is the probability that John wins the game in the hour?
2006-08-18 04:21:45 · 11 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
[(1/2)*20+(1/5)*40]/60= 3/10
2006-08-18 04:27:46 · answer #1 · answered by jimvalentinojr 6 · 1⤊ 0⤋
His chance of winning in the first 20 mins is 1/2.
His chance of winning in the last 40 mins is 1/5 but this is conditional upon him not winning previously. The chance of him not winning previously is 1/2, hence the conditional probability of him winning in the last 40 mins is 1/2 * 1/5 = 1/10.
His total chance of winning is therfore 1/2 + 1/10 = 6/10 = 3/5
2006-08-18 11:55:12 · answer #2 · answered by Stewart H 4 · 1⤊ 0⤋
Is the chance that the game is settled within the first 20 minutes 20/60 versus for 40/60 for the last forty minutes? In that case his chance of winning is 1/2*(20/60)+1/5*(40/60)=18/60 or 30%.
2006-08-18 11:29:53 · answer #3 · answered by helene_thygesen 4 · 0⤊ 0⤋
Your question is a little ambiguous. Are his chances "for each minute he plays?" Or are they per 20 minutes, then per 40 minutes?
Taking the second interpretation (because it's less arithmetic, and I'm lazy) let's examine the possibilities.
a)He wins in the first 20 minutes (1/2)
b)He does not win in the first 20 minutes (1/2), AND THEN he wins in the last 40 minutes (1/5). This is a compound--two events have to occur. Therefore the probablilty here is (1/2)*(1/5)=1/10.
The total probability of either (a) or (b) is the sum of the two disjoint probabilities. So the answer in this interpretation is 6/10.
2006-08-18 11:35:37 · answer #4 · answered by Benjamin N 4 · 2⤊ 0⤋
Given:
A = .5 = probability he wins in first 20 minutes
B = .2 = probability he wins in the second 40 minutes
C = ? =probability he wins in either period
We want to find the value of C
Assume:
D = probability that he wins in the first 20 minutes AND in the second 40 minutes = A * B = .1
If you draw a Venn diagram, you will see A and B overlap and that overlapping area is D. C is the area covered by A and B and D. To find this area, you can add A and B, but then you have D in there twice, so subtract one of the 'D's out.
C = A + B - D
C = .5 + .7 -.1
C = .6
C = 6/10 or 3/5 (expressed as a fraction to match the original description.)
2006-08-18 11:54:34 · answer #5 · answered by frd050101 2 · 0⤊ 0⤋
P(e)= Possible outcome/ total possible outcomes
P(winning the game)= P( first 20 minutes) + p(final 40 minutes)
P(e) = P(1/2) + p(1/5)
p(e) = 7/10 <+++++++++
2006-08-18 11:31:06 · answer #6 · answered by Carpe Diem (Seize The Day) 6 · 0⤊ 1⤋
(20/60)*(1/2) + (40/60)*(1/5) = .3 = 30%
2006-08-18 11:31:07 · answer #7 · answered by Will 6 · 0⤊ 0⤋
7/10 chance
2006-08-18 16:31:34 · answer #8 · answered by Anonymous · 0⤊ 0⤋
~24%
[.50 odds * (1/3) hours] + [.20 odds * (2/3) hours] = 24.44%
2006-08-18 11:28:01 · answer #9 · answered by HokiePaul 6 · 0⤊ 0⤋
no chance
2006-08-18 11:26:23 · answer #10 · answered by idontkno 7 · 0⤊ 0⤋