MATH SOLVE

4 months ago

Q:
# Matt flips $100$ coins. those that land heads, he sets aside. he then reflips the coins that landed tails, and again sets aside all those that land heads. finally, he flips a third time the coins that landed tails twice, and again sets aside all those that land heads.what is the expected number of coins matt sets aside?

Accepted Solution

A:

Let X be the random variable representing the number

of coins that Matt set aside after three flips.

[tex]P(X_1=n)=P(B(100,\frac{1}{2})=n),\text{ B is the binomial distribution.}\\P(X_2=m)=P(B(100-n,\frac{1}{2})=m)\\ P(X_3=l)=P(B(100-n-m,\frac{1}{2})=l)[/tex]

We have to run n from 0 to 100 and m from 0 to (100-n),

Find the binomial law, then compute the mean like this:

[tex] \frac{100-n-m}{2} [/tex]

Hope this gives a hint.

of coins that Matt set aside after three flips.

[tex]P(X_1=n)=P(B(100,\frac{1}{2})=n),\text{ B is the binomial distribution.}\\P(X_2=m)=P(B(100-n,\frac{1}{2})=m)\\ P(X_3=l)=P(B(100-n-m,\frac{1}{2})=l)[/tex]

We have to run n from 0 to 100 and m from 0 to (100-n),

Find the binomial law, then compute the mean like this:

[tex] \frac{100-n-m}{2} [/tex]

Hope this gives a hint.