Probability of independent studies question

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August 7th 2006, 09:32 AM
kwtolley

Probability of independent studies question

The probability that Andy studies math over break is 0.3 and the probability that Betty independently studies over break is 0.4. Find the probability Andy studies math over break and Betty doesn't.

P(B)=0.4 P(B)=1-0.4=0.6
P(A)=0.3 P(A)=1-0.3=0.7, Ok this is where I'm stuck do I add, Multi, or divide. List of possible answer are. My answer is 0.42 by going from my text book. Is this right. Thanks for any help given.

1. 0.12
2. 0.42
3. 0.18
4. 0.46
August 7th 2006, 09:56 AM
CaptainBlack

Quote:

Originally Posted by kwtolley

The probability that Andy studies math over break is 0.3 and the probability that Betty independently studies over break is 0.4. Find the probability Andy studies math over break and Betty doesn't.

P(B)=0.4 P(B)=1-0.4=0.6
P(A)=0.3 P(A)=1-0.3=0.7, Ok this is where I'm stuck do I add, Multi, or divide. List of possible answer are. My answer is 0.42 by going from my text book. Is this right. Thanks for any help given.

1. 0.12
2. 0.42
3. 0.18
4. 0.46

Yes these are independent events so their probabilities multiply.

That it cannot be the sum can be shown by supposing both events have
probability 0.7, clearly the probability of both occurring is <1, so it cannot be
the sum.

RonL
August 7th 2006, 10:07 AM
Quick

I'm just going to show the work (you can figure out why it works):

$P(A\not{B})=P(A)\times(1-P(B))$

substitute: $P(A\not{B})=0.3\times(1-0.4)$

subtract: $P(A\not{B})=0.3\times0.6$

multiply: $P(A\not{B})=0.18= 18\%$
August 7th 2006, 08:18 PM
kwtolley

thank you both

thanks quick for showing me the brake down.
August 7th 2006, 08:50 PM
CaptainBlack

Quote:

Originally Posted by kwtolley

thanks quick for showing me the brake down.

I for one am much happier answering you questions now that you are
indicating some of your thinking.

RonL