# Probability Questions

## A biased coin with probability p, 0 < p < 1, of heads is tossed until a head appears for the first time. If the probability that the number of tosses required is even, is 2/5, then p is equal to

Solution : Let the probability of getting a head be p and not getting a head be q. Since, head appears first time in an even throw 2 or 4 or 6. $$\therefore$$   $$2\over 5$$ = qp + $$q^3$$p + $$q^5$$p + …… $$\implies$$  $$2\over 5$$ = $$qp\over {1 – q^2}$$ Since q = 1- …

## 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

Solution : Let $$A_1$$, $$A_2$$, $$A_3$$ be the events of match winning in first, second and third matches respectively and whose probabilities are P($$A_1$$) = P($$A_2$$) = P($$A_2$$) = $$1\over 2$$ $$\therefore$$  Required Probability = P($$A_1$$)P($$A_2’$$)P($$A_3$$) + P($$A_1’$$)P($$A_2$$)P($$A_3$$) = $$({1\over 2})^3$$ + $$({1\over 2})^3$$  = $$1\over 8$$  + $$1\over 8$$ = $$1\over 4$$ Similar Questions …

## A fair die is tossed eight times. The probability that a third six is observed on the eight throw, is

Solution : Probability of getting success, p = $$1\over 6$$ and probability of failure, q = $$5\over 6$$ Now, we must get 2 sixes in seven throws, so probability is $$^7C_2$$$$({1\over 6})^2$$$$({5\over 6})^5$$ and the probability that 8th throw is $$1\over 6$$. $$\therefore$$   Required Probability = $$^7C_2$$$$({1\over 6})^2$$$$({5\over 6})^5$$$$1\over 6$$ = $$^7C_2\times 5^5\over {6^8}$$ Similar …

## If A and B are two mutually exclusive events, then

Question : If A and B are two mutually exclusive events, then (a) P(A) < P(B’) (b) P(A) > P(B’) (c) P(A) < P(B) (d) None of these Solution : Since, A and B are two mutually exclusive events. $$\therefore$$    A $$\cap$$ B = $$\phi$$ $$\implies$$ Either A $$\subseteq$$ B’ or B $$\subseteq$$ A’ …

## 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

Solution : The total number of ways in which numbers can be choosed = 25*25 = 625 The number of ways in which either players can choose same numbers = 25 $$\therefore$$ Probability that they win a prize = $$25\over 625$$ = $$1\over 25$$ Thus, the probability that they will not win a prize in …

## A problem in mathematics is given to three students A, B, C and their respective probability of solving the problem is $$1\over 2$$, $$1\over 3$$ and $$1\over 4$$. Probability that the problem is solved is

Solution : Since the probability of solving the problem by A, B and C is $$1\over 2$$, $$1\over 3$$ and $$1\over 4$$ respectively. $$\therefore$$  Probability that problem is not solved is = P(A’)P(B’)P(C’) = ($$1 – {1\over 2}$$)($$1 – {1\over 3}$$)($$1 – {1\over 4}$$) = $$1\over 2$$$$2\over 3$$$$3\over 4$$ = $$1\over 4$$ = Hence, the …

## Five horses are in a race. Mr A selects two of the horses at random and bets on them. The probability that Mr A selected the winning horse is

Solution : The probability that Mr A selected the lossing horse = $$4\over 5$$ $$\times$$ $$3\over 4$$ = $$3\over 5$$ The probability that Mr A selected the winning horse = 1 – $$3\over 5$$ = $$2\over 5$$ Similar Questions Consider rolling two dice once. Calculate the probability of rolling a sum of 5 or an …

## Consider rolling two dice once. Calculate the probability of rolling a sum of 5 or an even sum.

Solution : Sample space of rolling two dice = 36 Now, Event of sum of 5 = { (1,4) (2,3) (3,2) (4,1) } = 4 $$\implies$$ probability of getting sum of 5 is 4/36 = 1/9 Now, Event of even sum = { (1,1) (1,3) (1,5) (2,2) (2,4) (2,6) (3,1) (3,3) (3,5) (4,2) (4,4) (4,6) …

## The probability that A speaks truth is 4/5 while this probability for B is 3/4. The probability that they contradict each other when asked to speak on a fact is

Solution : Given, probabilities of speaking truth are P(A) = $$4\over 5$$ and P(B) = $$3\over 4$$ And their corresponding probabilities of not speaking truth are P(A’) = $$1\over 5$$  and P(B’) = $$1\over 4$$ The probability that they contradict each other = P(A).P(B’) + P(B).P(A’) = $$4\over 5$$ $$\times$$ $$1\over 4$$ + $$1\over 5$$ …

## 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

Solution : $$\therefore$$  Total number of cases = $$3^3$$ Now, favourable cases = 3(as either all has applied for house 1 or 2 or 3) $$\therefore$$ Required Probability = $$3\over 3^3$$ = $$1\over 9$$ Similar Questions The probability of India winning a test match against the west indies is 1/2 assuming independence from match to …