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%

Chuckrox

Wouldn’t it just be P(not A) and P(not B) = 0.4 x 0.65 = 0.26

pacmandefense

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%

liquidity

Basically just multiply probabilities together.  I use this whenever I’m tempted to play the lottery.

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???

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

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.

ramzes

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

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?

Zesty

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