There are some Red and Blue balls in a bag of 21 balls and the probability of picking two red balls out in a row is 1/14. How many red balls are there?

Question

Please help.... I have no idea how to do this... i would appreciate any help with the working out and the answer. Thank you

Answer 1

The probability of picking the first red ball out of the bag is (number of red balls)/(total number of balls).  Since we don't know how many reds there are, we will use the variable R to represent the number of red balls.  So the probability of the first red ball is R/21.

 

The probability of picking the second red ball out of the bag is (number of remaining red balls)/(total number of balls left in the bag). So the probability of the second red ball is (R-1)/20.

 

When performing multiple events in a row, we multiply the probabilities of each event.  This allows us to write the equation:

(R/21)*((R-1)/20)=(1/14)

 

Solving:

(R²-R)/420 = (1/14)

Cross-multiplying:

14(R²-R) = 420

 

Continuing to solve:

14R²-14R = 420 

14R²-14R-420 = 0

R²-R-30 = 0

(R-6)(R+5) = 0

 

Ignoring the negative solution, we get R = 6.  So there are 6 red balls in the bag.