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Sagot :
Answer:
5
Step-by-step explanation:
The winnings are in G.P. : 1, 2, 4, ..... till 10 toss.
[tex]$a_n = 1 \times 2^{n-1}\ \ \ \forall \ n = 1,2,3,4,....,10$[/tex]
[tex]$a_n$[/tex] denotes the winnings on the [tex]$n^{th}$[/tex] toss.
The probability of earning amount [tex]$a_n$[/tex] on the [tex]$n^{th}$[/tex] toss is = [tex]$\left(\frac{1}{2}\right)^n$[/tex]
∴ [tex]$E(X) = \sum_{n=1}^{10} \ a_n \times \left(\frac{1}{2}\right)^n $[/tex]
[tex]$=\sum_{n=1}^{10} \ 1 \times \frac{2^{n-1}}{2^n} $[/tex]
[tex]$=\sum_{n=1}^{10} \ \frac{1}{2}$[/tex]
Sum of the 1st n terms of the A.P. is :
[tex]$=\frac{n}{2}[2a+(n-1)d] $[/tex]
[tex]$=\frac{10}{2}[2\times \frac{1}{2}+(10-1)\times 0] $[/tex]
= 5
Therefore, E(X) = 5
Hence the expected value of the game is 5
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