Probability question ... please explain television, cable, following, question, neither

bchadwick

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%

sabina

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.

simeezee

why is it not correct to calculate p(A andB)= 0.21 and then p(non(Aand B)= 1 - 0.21 = 0.79?

Wasteoftime

Sorry for the double comment, I am still new here.

Zesty

Sorry for the double comment, I am still new here.

chunty

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?

ramzes

Isn’t it incorrect to use the joint probablitiy there?

willsucceed

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.

prav_Cfa7

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

Sunshine4ever

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???