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
Hi everyone,
Could someone please help me on this conditional probablity problem?
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
Thank you very much
I understand "d" now, but I still don't understand "a".
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"
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
Thank you