Probablility PS

[theachiever]

Gender Under 2.0 2.0-3.0 Over 3.0

Male 0.05 0.25 0.10

Female 0.10 0.30 0.20

"¢If the student selected is female, what is the probability that her GPA is between 2.0 and 3.0?
"¢If the GPA of the student selected is over 3.0, what is the probability that the student is male?
"¢What is the probability that the student selected is female or has a GPA under 2.0 or both?
"¢ Is GPA independent of gender?

[faraz_jeddah]

I believe that the question is in a table format (3x3)


if the student selected is female, what is the probability that her GPA is between 2.0 and 3.0?
Answer - 0.3

If the GPA of the student selected is over 3.0, what is the probability that the student is male?
Answer - 0.1 / (0.1 + 0.2) = 1/3 = 0.33

What is the probability that the student selected is female or has a GPA under 2.0 or both?

Answer = 0.6 + 0.15 + 0.1 = 0.85

Is GPA independent of gender?
Yes ?

[jitsy]

This involves conditional probability. The probability of A conditioned on B, denoted P(A|B), is equal to P(AB)/P(B). The division provides that the probabilities of all outcomes within B will sum to 1. Conditioning restricts the sample space to those outcomes which are in the set being conditioned on (in this case B).

Answering question 1

1) If the student selected is female, what is the probability that her GPA is between 2.0 and 3.0?

Here
P(A)= 0.3/0.6 = 0.5
P(B)= 0.6/1.0 = 0.6
P(A|B)= P(AB)/P(B)= P(A)*P(B)/P(B) = 0.5*0.6/0.6= 0.5

[theachiever]

Thanks for the solutions guys.I appreciate the help

Experts can you please give your inputs about the problem if you can guide me about it,it would be helpful.

[Matt@VeritasPrep]

Let's do it!

The first question is asking what percentage of the females have a GPA of 3.0, so our probability is:

# of 3.0 females
----------------
Total # of females

= .3/.6 = .5 = 50%

The second question is similar

# of males over 3.0
-------------------
Total # of people over 3.0

= .1/.3 = 1/3 = 33.3..%

The third one is tricky: for the odds of AT LEAST ONE of two groups, we do (Odds A) + (Odds B) - (Odds of A + B).

So ...

Odds Female = .6/1 = 60%
Odds Under 2.0 = .15/1 = 15%
Odds Both = .1/1 = 10%

So 60% + 15% - 10% = 65%

The other way to find the odds of AT LEAST ONE of two things is to do 1 - (Odds of neither). So the odds that someone is NOT a female or a person with a GPA under 2.0 is .35 (Guys with 2.0-3.0 or Guys with 3.0+). So 1 - .35 = .65, or 65%.

[jitsy]

Matt,

Could you please help me with the last part of the (Yes/No) question and also explaining the reasoning. Thank you. I think it is the same as the last part in the question below. Just need some guidance in answering questions like these.

On 43percent of days, the market actually goes up. On 12 percent of days, Peter predicts that the market will go up and it does. If we pick a day at random, what is the probability that:
"¢Peter will predict that the market will go up, or it will actually go up, or both?
"¢The market will actually go up given that Peter has predicted it will?
"¢Does the Stock market moves up independently of Peter's Prediction?

Let's do it!

The first question is asking what peurcentage of the females have a GPA of 3.0, so our probability is:

# of 3.0 females
----------------
Total # of females

= .3/.6 = .5 = 50%

The second question is similar

# of males over 3.0
-------------------
Total # of people over 3.0

= .1/.3 = 1/3 = 33.3..%

The third one is tricky: for the odds of AT LEAST ONE of two groups, we do (Odds A) + (Odds B) - (Odds of A + B).

So ...

Odds Female = .6/1 = 60%
Odds Under 2.0 = .15/1 = 15%
Odds Both = .1/1 = 10%

So 60% + 15% - 10% = 65%

The other way to find the odds of AT LEAST ONE of two things is to do 1 - (Odds of neither). So the odds that someone is NOT a female or a person with a GPA under 2.0 is .35 (Guys with 2.0-3.0 or Guys with 3.0+). So 1 - .35 = .65, or 65%.