Question

A box contains 4 bad and 6 good tubes. Three are drawn out together. One of them
is tested and found to be good. What is the probability that third one is bad?

Answer 1

Answer:

(\frac{5}{9}\)

Step-by-step explanation:

Let A= one of the tubes drawn is good and

B=The other tube is good .

P(A∩B) =P(both tubes drawn are good)

6C210C2=13

knowing that one tube is good,the conditional probability that the other tube is also good is required i.e P(B/A)is required.

P(B/A)=P(A∩B)P(A)