Short Answer TypeA and B toss a coin alternatively till one of them gets a head and wins the game. If A starts the game, find the probability of his winning at his third toss.
Long Answer TypeShort Answer TypeThere are three urns A, B and C. Urn A contains 4 white balls and 5 blue balls. Urn B contains 4 white balls and 3 blue balls. Urn C contains 2 white balls and 4 blue balls. One half is drawn from each of these urns. What is the probability that out of these three balls drawn, two are white balls and one is a blue ball?
Let A be the event that the construction job will be completed on time, and B be the event that there will be a strike.
Now    P(B) = 0.65
P(no strike) = P(B’) = 1 - P(B) = 1 - 0.65 = 0.35
P(A | B) = 0.32, P(A | B') = 0.80
Since events B and B' form a partition of the sample space S,
∴  By theorem on total probability, we have
P(A) = P(B) P(A | B) + P(B') P(A | B’)
= 0.65 × 0.32 + 0.35 × 0.8
= 0.208 + 0.28 = 0.488
∴  the probability that the construction job will be completed in time is 0.488.
An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?
Switch
