标题: Probability question ... please explain [打印本页] 作者: bchadwick 时间: 2013-4-28 07:07 标题: Probability question ... please explain
Consider the following two independent events and corresponding probabilities: Event A) The probability that the auto demand will rise more than 5% during the coming year is 60%. Event B) The probability that the demand for cable television will rise more than 10% is 35%. The probability that neither events will occur is:
Select one:
a. 74%
b. 21%
c. 26%作者: Chuckrox 时间: 2013-4-28 07:08
Wouldn’t it just be P(not A) and P(not B) = 0.4 x 0.65 = 0.26作者: pacmandefense 时间: 2013-4-28 07:08
My answer would be c. 26%.
My approach to the solution:
The probability of auto demand rising more than 5% during the coming year is 60%.
Therefore, the probability of auto demand NOT rising more than 5% during the coming year is = (100% - 60%) = 40% = 0.4
The probability that the demand for cable television will rise more than 10% is 35%.
Therefore, the probability that the demand for cable television will NOT rise more than 10% is = (100% - 35%) = 65% = 0.65
The probability that both these independent events will NOT occur = P(auto demand not rising) X P (cable demand not rising) = 0.4 X 0.65 = 0.26 = 26%作者: liquidity 时间: 2013-4-28 07:08
Basically just multiply probabilities together. I use this whenever I’m tempted to play the lottery.作者: Sunshine4ever 时间: 2013-4-28 07:09
Yes, yours is one of the right ways, but costs you more time.
To be simple:
P(~A) = 0.4 and P(~B) = 0.65
Therefore, P(~A and ~B) = 0.4 * 0.65 = 0.26
By the way, you made two typos: 1) 0.25 but not 26 on 2nd line, and, 2) 0.6 + 0.35 instead of 0.6 + 0.36 on 5th line.
dsjn358 wrote:
I’m not sure if this is right but:
I gota answer C- 0.25
P(A: auto 5%) = 0.6
P(B: cable10%)=0.35
P(A: auto 5% or B: cable10%) = 0.6 + 0.36 - (0.6 x 0.35) = 0.95 - 0.21 = 0.74
P(NOT A: auto 5% or B: cable10%) = 1-0.74 = 0.26
is this the right way to think of this problem???作者: prav_Cfa7 时间: 2013-4-28 07:09
EddieChen wrote:
Yes, yours is one of the right ways, but costs you more time.
To be simple:
P(~A) = 0.4 and P(~B) = 0.65
Therefore, P(~A and ~B) = 0.4 * 0.65 = 0.26
By the way, you made two typos: 1) 0.25 but not 26 on 2nd line, and, 2) 0.6 + 0.35 instead of 0.6 + 0.36 on 5th line.
dsjn358 wrote:
I’m not sure if this is right but:
I gota answer C- 0.25
P(A: auto 5%) = 0.6
P(B: cable10%)=0.35
P(A: auto 5% or B: cable10%) = 0.6 + 0.36 - (0.6 x 0.35) = 0.95 - 0.21 = 0.74
P(NOT A: auto 5% or B: cable10%) = 1-0.74 = 0.26
is this the right way to think of this problem???
thanks mate, the other way is much easier!
oops yes my bad,
answer C = 0.26
and yes, P(A: auto 5% or B: cable10%)= 0.6+0.35- (0.6 x 0.35) = 0.95 - 0.21 = 0.74作者: willsucceed 时间: 2013-4-28 07:09
Just feel the intuition behind it before you answer the question. First calculate the probablitites that those events won’t occur. Those are 40% and 65% for A complement and B complement respectively. Then, see if there is a joint probability. Since these are independant events, there isn’t a joint probability. Therefore the answer is just the multiplication of the two probabilites of the complement of those events hapening. Therefore, the answer if 26% or C.作者: ramzes 时间: 2013-4-28 07:09
Isn’t it incorrect to use the joint probablitiy there?作者: chunty 时间: 2013-4-28 07:10
dsjn358 wrote:
EddieChen wrote:
Yes, yours is one of the right ways, but costs you more time.
To be simple:
P(~A) = 0.4 and P(~B) = 0.65
Therefore, P(~A and ~B) = 0.4 * 0.65 = 0.26
By the way, you made two typos: 1) 0.25 but not 26 on 2nd line, and, 2) 0.6 + 0.35 instead of 0.6 + 0.36 on 5th line.
dsjn358 wrote:
I’m not sure if this is right but:
I gota answer C- 0.25
P(A: auto 5%) = 0.6
P(B: cable10%)=0.35
P(A: auto 5% or B: cable10%) = 0.6 + 0.36 - (0.6 x 0.35) = 0.95 - 0.21 = 0.74
P(NOT A: auto 5% or B: cable10%) = 1-0.74 = 0.26
is this the right way to think of this problem???
[snip]
thanks mate, the other way is much easier!
oops yes my bad,
answer C = 0.26
and yes, P(A: auto 5% or B: cable10%)= 0.6+0.35- (0.6 x 0.35) = 0.95 - 0.21 = 0.74
Isn’t it incorrect to think that there is a joint probability there?作者: Zesty 时间: 2013-4-28 07:10
Sorry for the double comment, I am still new here.作者: Wasteoftime 时间: 2013-4-28 07:10
Sorry for the double comment, I am still new here.作者: simeezee 时间: 2013-4-28 07:11
why is it not correct to calculate p(A andB)= 0.21 and then p(non(Aand B)= 1 - 0.21 = 0.79?作者: sabina 时间: 2013-4-28 07:11
ana_georgiana90 wrote:
why is it not correct to calculate p(A andB)= 0.21 and then p(non(Aand B)= 1 - 0.21 = 0.79?
In this case you are looking for P(not(AorB)) not p(not(AB)). Therefore, once you get P(AandB) you need to use addition rule P(A or B)= P(A) + P(B)-P(AandB)= 0.60+0.35-0.21= 0.74. P(AorB)= 0.74 so P(not(AorB) = .026.
