An urn B1 contains 2 white and 3 black balls and another urn B2 contains 3 white and 4 black balls. One urn is selected at random and a ball is drawn from it. If the ball drawn is found black, find the probability that the urn chosen was B1.

P(B1|Black) = = 21/41 = 0.512

Step-by-step explanation:

In this question, we will use the conditional probability formula that is, for two events A and B, the probability that event A occurs given than event B has already occured is:

P(A|B) = P(A∩B)/P(B)

Here, we need to find the probability that the ball is from urn B1 given that it is black. i.e.

P(B1|Black) = P(B1∩Black) / P(Black)

P(B1∩Black) is the probability that the ball is chosen from B1 and it is black. The number of black balls in urn B1 is 3 and the total number of balls in this urn is 5. The probability of choosing either of the urns is 1/2.

So, P(B1∩Black) = (1/2) (3/5)

     P(B1∩Black) = 3/10

P(Black) is the probability of selecting a black ball. this can be from either of the urns B1 and B2. So, we can calculate this probability as:

P(Black) = P(Black in B1) + P(Black in B2)

             = (1/2)(3/5) + (1/2)(4/7)

P(Black) = 41/70

P(B1|Black) = P(B1∩Black) / P(Black)

                  = (3/10) / (41/70)

P(B1|Black) = 21/41 = 0.512

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b) 62.47% of the students are between 14 and 18 years old.

step-by-step explanation:

mean = 15 and standard deviation = 2

student between 14 to 18

formula used z-score

for age 14 , z score

using z-table , p(z> -0.5)=0.3085

for age 18 , z score

using z-table , p(z< 1.5)=0.9332

p(-0.5< z< 1.5)=0.9332-0.3085 = 0.6247

thus, 62.47% of the students are between 14 and 18 years old.

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