A bag contains 4 red and 4 blue balls. Four balls are drawn one by one from the bag, then find the probability that the drawn balls are in alternate color.

Solution :

\(E_1\) : Event that first drawn ball is red, second is blue and so on.

\(E_2\) : Event that first drawn ball is blue, second is red and so on.

\(\therefore\)  P(\(E_1\)) = \(4\over 8\) \(\times\) \(4\over 7\) \(\times\) \(3\over 6\) \(\times\) \(3\over 5\) and

\(\therefore\)  P(\(E_2\)) = \(4\over 8\) \(\times\) \(4\over 7\) \(\times\) \(3\over 6\) \(\times\) \(3\over 5\)

P(E) = P(\(E_1\)) + P(\(E_2\)) = 2 \(\times\) \(4\over 8\) \(\times\) \(4\over 7\) \(\times\) \(3\over 6\) \(\times\) \(3\over 5\) = \(6\over 35\)

---

Similar Questions

The probability of India winning a test match against the west indies is 1/2 assuming independence from match to match. The probability that in a match series India’s second win occurs at the third test is

Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that three apply for the same house is

Find the number of ways of dividing 52 cards among 4 players equally such that each gets exactly one Ace.

A and B play a game, where each is asked to select a number from 1 to 25. If the two numbers match, both of them win a prize. The probability that they will not win a prize in a single trial is

A die is thrown. Let A be the event that the number obtained is greater than 3. Let B be the event that the number obtained is less than 5. Then, \(P(A \cup B)\) is