Question

QUESTION 40. An urn contains equal number of green and red balls.
Suppose you are playing the following game. You draw one ball at random
from the urn and note its colour. The ball is then placed back in the urn,
and the selection process is repeated. Each time a green ball is picked you
get 1 Rupee. The first time you pick a red ball, you pay 1 Rupee and the
game ends. Your expected income from this game is​

Answer 1

Answer:

Step-by-step explanation:

The expected value of our income is given by

()=∑

over all outcomes , where is the income and the probability of that outcome.

Let the th outcome be drawing times. Then the first −1 drawings will have been green, and the last drawing will be red, so we get income =−2.

The probability of the th outcome is (first −1 balls are green)∗(last ball is red). Since we replace the ball, these events are independent, so =(12)−1∗12=(12).

Now we can sum over all outcomes to find the expected value:

()=∑=1∞[(−2)(12)]=∑=1∞−22

This converges by the ratio test, so we can do some manipulation:

2()=∑=1∞−22−1=∑=0∞−12=−1+∑=1∞−12

And then subtract:

2()−()=(−1+∑=1∞−12)−(∑=1∞−22)=−1+∑=1∞(−1)−(−2)2=−1+∑=1∞12=−1+1=0=(