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# need help on probability:

A study involved married husbands and wives
in prime time 60% husbands are watching TV, when husband is watching, 30% of the time wife is also watching.When husband is not watching, 10 % of the time wife is watching TV

1.Probabilty that wife is watching Tv while husband is also watching TV

2.Probability that wife is watching TV during Prime time:

Please explain how to compute this rounding to three decimal places

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Table_Data,     B,   B`
A 10,  40
A' 40, 20
P(A|B)
P(A|B')
P(A'|B')

Are events A and B independent?

### 2 Answers by Expert Tutors

Kay G. | ~20 Years Accounting Tutoring Experience~20 Years Accounting Tutoring Experience
4.9 4.9 (32 lesson ratings) (32)
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This is a little tricky.  If you want to use the chart, you need to get it corrected and understand how to use it.

Start with the numbers you know, and don't start with the conditions.  The conditions don't work how you're putting them into the chart.  (I think you're getting turned around as well.)

On the chart that figure looks like:
B       B'
A
A'   __________
.60              1.00

Notice where that's going.  The B column total is the probability of hubby watching TV.  The B' column total is the probability he's not watching TV.  The total across the bottom must = 1.00.  If the probability that hubby is watching TV is .60, then what is the probability he's not watching?  If you also total the A rows (out to the right), and then total those down, that also = 1.00.

The space where A and B cross is not a conditional - it's an intersection.  That would be A∩B.  We don't know that number, but we can solve it.  But it's not .30.

If you want the intersection, you need to solve for it:

P(A|B) = [P(A∩B)] / [P (B)]

(You should know this equation.)

You have P(A|B) as given in the problem.  You have P(B) as given in the problem.  Plug in and solve for P(A∩B).  That is what goes on the chart, where A and B cross, because that's the intersection.

(You should now be able to fill in where A' and B cross if you like, since that column has to add to .60.  But it's not necessary to solve the problem, but good practice. :-))

Now, if you've already filled in the total on the B' column, then we can do a similar thing to find the cross between A and B'.  That is the intersection of wife watching & husband not watching.  That again is not the 10% of the condition - it's the intersection:  P (A∩B').

P (A|B') = [P(A∩B'] / [P(B')]

Same equation I used above.  The conditional equation is the intersection of the two, divided by the condition or the second item.  That always works.

You were given P (A|B') in the problem.  You should have solved for the P(B') at the bottom of the B' column on the chart.  Plug in and solve for P(A∩B').

The answer to that goes in the space that crosses A and B' in the chart, cause that's the intersection.

You should now have both numbers for A:  where it crosses B and where it crosses B'.  That total is the probability of the wife watching.  That adds across.  (You can actually fill in the entire chart from here, cause the columns are totaled, the rows are totaled, and the total of both those should add up to 1.00 in the bottom right corner.)

Independence:
They are independent if P(A) = P(A|B)
If you have now solved for P(A), you can check if this is equal.

The trick is understanding how conditionals work, and in understanding that the 30% is out of the 60%, not out of the total.  That's because we've dumped out the other 40% of the husbands and they don't exist in that conditional probability.  So out of the 60%, 30% of just those are watching.

Just see what you can do with this.  You can post anything you come up with.

Meena S. | Patient & knowledgeable Mathematics & Statistics (SPSS) TutorPatient & knowledgeable Mathematics & St...
4.5 4.5 (57 lesson ratings) (57)
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Hi Rekha,

Let Husband is watching TV in prime time be A.
Wife is watching TV in Prime time be B.
60% Husbands are watching TV so
P(A)= 0.60
P ( Husband is not watching TV)= P(A')= 1-0.60=0.40

when husband is watching, 30% of the time wife is also watching.

1) Probabilty that wife is watching Tv while husband is also watching TV= 0.30.

2) Probability that wife is watching TV during Prime time= P(B)=
P( 30% of time Wife is watching TV when Husband is watching) + P( When husband is not watching, 10 % of the time wife is watching TV)
=0.30+0.040= 0.34.
P(B')= 1- 0.34=0.66.

events A & B are independent so
p(A/B)= P(A)=0.60
P(A/B')= P(A)P(B)/P(B')=(0.60)(0.34)/0.66=0.3091.
P(A'/B')=P(A')P(B')/P(B')= 0.40 * 0.66/0.66=0.40= P(A')

Meena from Strongsville,Oh

You got the .04 by taking 10% of .40.  i.e. P(wife|husband') * P(husband').
But you left the .30 alone, rather than taking P(wife|husband) * P(husband).
In other words, it's inconsistent.

The problem says this:
"when husband is watching, 30% of the time wife is also watching.When husband is not watching, 10 % of the time wife is watching TV"
but you have treated one as a condition and one as not.

You've not shown where that .04 came from.

You've determine they're independent but have not shown where you came to this conclusion.