Probability Questions

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- …

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 Read More »

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 …

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 Read More »

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 …

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

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’ …

If A and B are two mutually exclusive events, then Read More »

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 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 Read More »

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 …

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 Read More »

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 …

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 Read More »

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) …

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

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’) = \(4\over 5\) \(\times\) \(1\over 4\) + \(1\over 5\) …

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 Read More »

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 …

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 Read More »