Suppose that a box contains 5 red balls and ten blue balls. Further assume that we draw balls at random with replacement until we see three red balls. What is the probablility that we will see 2 blue balls before seeing 3 red balls?

8/81

Step-by-step explanation:

Total number of balls = 10 + 5 = 15balls

Since it is with replacement, then

Probability of picking a red ball = 5/15

Probability of picking a blue ball = 10/15

The order of picking in which we can have 2 blue balls before 3 red balls becomes:

= [BBRRR], [BRBRR], [BRRBR], [RBBRR], [RBRBR], [RRBBR]

For the first order of occurrence = [10/15 * 10/15 * 5/15 * 5/15 * 5*15] = 12500/759375 = 4/243.

Since this is the same value we will get for all the possible order of occurrence stated, the we multiply this answer by the number of occurrence.

Hence, probability of getting 2 blue balls before 3 red balls = (4/243) × 6

=24/243

=8/81

his chances of getting a coin of higher value is 4/19

step-by-step explanation: