# conditional probablity

• Sep 23rd 2007, 03:54 PM
chocolatelover
conditional probablity
Hi everyone,

Given:
Coin tossing
A balanced dime is tossed twice. The four possible equaly likely outcomes are HH,HT, TH, TT

Let:

A=event the first toss is heads
B=event the second toss is heads
C=event at least one toss is heads

Determine the following probabilities and express your results in words. Compute the conditional probabilities directly; do not use the conditional probability rule

a. P(B)

Would it be 3/4???????

b. P(B|A)

Would it be 1/3????????

c. P(B|C)

Would this also be 1/3???????

d. P(c)

1/3????

e. P(C|A)

f. P(|not B))

I really don't understand this. Could someone please explain this to me so that I can do the rest that are like this?

any help would be greatly appreciated

Thank you very much
• Sep 23rd 2007, 04:17 PM
MathGuru
For a)

The possibilities are HH, HT, TH, TT

so the probability that the second toss is heads is 2 out of the four possibilities. It makes sense because HT and TT are the only two that satisfy this condition. Therefore, 1/2
• Sep 23rd 2007, 04:19 PM
MathGuru
For d)

3 out of the four possibilities have a H in them HT, TH, HH

• Sep 23rd 2007, 04:27 PM
chocolatelover
Thank you very much

I understand "d" now, but I still don't understand "a".
Quote:

For a)

The possibilities are HH, HT, TH, TT

so the probability that the second toss is heads is 2 out of the four possibilities. It makes sense because HT and TT are the only two that satisfy this condition. Therefore, 1/2
How do you know that "HH" and "TT" are the only two that satisfy this condition? Wouldn't it be "HH" "HT" and "TH"? because they are the ones that have a "heads"

Thank you