Question

a bag contains 4 red marbles and 1 black amrble. two people, A and B, take turns drawing a marble from the bag without replacement. the first person to draw a black marble wins. What is the probability that the first person who draws wins

Answer 1

Answer:

Chances of winning = $\frac {1}{10}$ or 0.1

Step-by-step explanation:

According to the question, there may be two conditions:

1- Either A or B may draw a black marble first.

2- A or B both can draw the marble simultaneously.

This conclude that both of them have equal chances of getting the marble.

In both the conditions presented above, chances of drawing black marble by A or B is 50% or $\frac {1}{2}$.

Since, probability of occurrence of an event P = No. of possible outcomes/ total number of outcomes (exhaustive events).

Here, total possible outcomes = 2 (Either A or B wins)

So, Prob. that A wins = $\frac {1}{2}$

& Prob. that B wins = $\frac {1}{2}$

Now, let us talk about marbles:

We have 4 red marbles & 1 black marble.

In this case, total possible outcomes = 4+1 = 5

Prob. of a red marble = $\frac {4}{5}$

Prob. of a black marble = $\frac{1}{5}$

Thus, Prob. that A draws the black marble = $\frac{{1}{5} \times {1}{2}}$

                                                                      = $\frac {1}{10}$

and Prob. that B draws the black marble = $\frac{{1}{5} \times {1}{2}}$

                                                                      = $\frac {1}{10}$

Therefore, whether A or B , any of the persons may first draw the black marble, the prob. of winning = $\frac {1}{10}$