Question 301258
<pre><font size = 4 color = "indigo"><b>
Mean number of heads = 0*P(0 heads) + 1*P(1 head) + 2*P(2 heads) + 3*P(3 heads).

We don't need the probability of 0 heads since that will be multiplied
by 0. So:

Mean number of heads = 1*P(1 head) + 2*P(2 heads) + 3*P(3 heads).

The probability of getting x heads out of n tosses of a fair coin is

{{{nCx*(1/2)^x*(1/2)^(n-x)=nCx*(1/2)^n}}}

Use that formula to fill in this table of probabilities:

                   P(1 head)    P(2 heads)    P(3 heads)
ball#1(1 toss)         1/2           0             0      
ball#2(2 tosses)       1/2          1/4            0
ball#3(3 tosses)       3/8          3/8           1/8 

Now we need to get P(1 head), P(2 heads), P(3(heads)

P(1 head) = P[(ball#1 AND 1 head) OR (ball#2 AND 1 head) OR (ball#3 AND 1 head)]

Using the rule of "AND implies multiplication" and "OR implies addition":

P(1 head) = {{{(1/3)*(1/2)+(1/3)*(1/2)+(1/3)*(3/8)=1/6+1/6+1/8=11/24}}}

P(2 heads) = P[(ball#2 AND 2 heads) OR (ball#3 AND 2 heads)] = {{{(1/3)*(1/4)+(1/3)*(3/8)=1/12+1/8=5/24}}}

P(3 heads) = P[ball#3 AND 3 heads] = {{{(1/3)*(1/8)=1/24}}}

So,

Mean number of heads = 1*P(1 head) + 2*P(2 heads) + 3*P(3 heads) = {{{1*(11/24) + 2*(5/24) + 3*(1/24)=11/24+10/24+3/24=24/24=1}}}.

So, surprisingly, the mean number of heads is 1 head!!!

Edwin</pre>