Question

Suppose you toss a coin and will win $1 if it comes up heads. If it comes up tails, you toss again. This time you will receive $2 if it comes up heads. If it comes up tails, toss again. This time you will receive $4 if it is heads. Continue in this fashion for a total of 10 flips of the coin, after which you receive nothing if it comes up tails. What is the mathematical expectation for this game?

Answer 1

Answer:

5

Step-by-step explanation:

The winnings are in G.P. : 1, 2, 4, ..... till 10 toss.

$a_n = 1 \times 2^{n-1}\ \ \ \forall \ n = 1,2,3,4,....,10$

$a_n$ denotes the winnings on the $n^{th}$ toss.

The probability of earning amount $a_n$ on the $n^{th}$ toss is = $\left(\frac{1}{2}\right)^n$

∴ $E(X) = \sum_{n=1}^{10} \ a_n \times \left(\frac{1}{2}\right)^n$

            $=\sum_{n=1}^{10} \ 1 \times \frac{2^{n-1}}{2^n}$

           $=\sum_{n=1}^{10} \ \frac{1}{2}$

Sum of the 1st n terms of the A.P. is :

$=\frac{n}{2}[2a+(n-1)d]$

$=\frac{10}{2}[2\times \frac{1}{2}+(10-1)\times 0]$

= 5

Therefore, E(X) = 5

Hence the expected value of the game is 5