At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the **expert** answer you'll need to create a **free** account at **Brainly**

sample space is now
HHH
HHT
HT

Nice @ satellite 73

Not the answer you are looking for?

Search for more explanations.

in two out of the 3 cases it is thrown three times, unless i misinterpreted the question

ty

given answer is actually 1/2

no
\[P(A\cap B)\neq \frac{1}{2}\]

{T, HT, HHT, HHH} only 4 possiblities

exactly what you wrote yes

\[\frac{3}{4}\div \frac{1}{2}>1\]

well i made a mistake of course, (not in the answer, in the computation)

P(A) is correct, how to get P(A∩B)?

ok i see the problem

i am sorry i am an idiot this morning, so lets go slow

that is, each does not occur with probability 1/4

\[P(T)=\frac{1}{2}, P(HT)=\frac{1}{4},P(HHH)=\frac{1}{8},P(HHH)=\frac{1}{8}\]

ah, yeah..

ok now i understood lol

so i was completely wrong, and i apologize. should have a nice cup of coffee

ok so now let us finish the problem.

so \[ P(A \cap B)=\frac{1}{8}+\frac{1}{8}=\frac{1}{4}\]

|dw:1326555438102:dw|

we know
\[A\cap B = \{HHT, HHH\}\] and so what you said,
\[P(A\cap B)=\frac{1}{4}\]

sorry again for the error and taking up so much time being wrong

Not the answer you are looking for?

Search for more explanations.