Question

A & b are alternately picking balls from a bag without replacement. The bag has k black balls and 1 red ball. Winner is the one who picks the red ball. Who is more likely to win, the on who starts first, or second?

Answer 1

Answer:

The one who starts first mans a

Answer 2

Given a bag containing k black and one red balls, Find who is more likely to win if a red ball drawn is considered winning.

Explanation:

• From the concept of empirical probability we know that the probability of an event 'E' to occur is calculated by dividing the number of outcomes where 'E' has occurred by the total number of possible outcomes.  $P(E)=\frac{n(E)}{N}$  
• Here the event 'E' is that a red ball is drawn. Hence, $n(E)=1$  
• Now in the case of picking without replacement the total number of balls will decrease by one for consecutive draws.
• Hence, the total number of balls for first pick is $N=k+1$ and for second pick is $N'=k$
• Probabilities of drawing a red ball for each is given by, $P_1(E)=\frac{1}{k+1} \ \ \ \ \ \ \ ->P_2(E)=\frac{1}{k}$
• Since k is a positive , $P_2(E)>P_1(E)$
• Hence the person who starts second is more likely to win.