Probability Questions

Solution : Required Probability = P(X = 0) + P(X = 1) = \(e^{-5}\over 0!\).\(5^0\) + \(e^{-5}\over 1!\).\(5^1\) = \(e^{-5}\) + 5\(e^{-5}\) = \(6\over {e^5}\) 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 …

At a telephone enquiry system, the number of phone calls regarding relevant enquiry follow. Poisson distribution with an average of 5 phone calls during 10 min time intervals. The probability that there is almost one phone call during a 10 min time period is Read More »

Solution : Let the events, A = Ist aeroplane hit the target B = 2nd aeroplane hit the target And their corresponding probabilities are P(A) = 0.3 and P(B) = 0.2 \(\implies\) P(A’) = 0.7 and P(B’) = 0.8 \(\therefore\)  Required Probability = P(A’)P(B) + P(A’)P(B’)P(A’)P(B) + ……. = (0.7)(0.2) + (0.7)(0.8)(0.7)(0.2) + ……. = …

Two aeroplanes I and II bomb a target in succession. The probabilities of I and II scoring a hit correctly are 0.3 and 0.2, respectively. The second plane will bomb only if the first misses the target. The probability that the target is hit by the second plane, is Read More »

Solution : Probability of getting score 9 in a single throw = \(4\over 36\) = \(1\over 9\) Probability of getting score 9 exactly in double throw = \(^3C_2\) \(\times\) \(({1\over 9})^2\) \(\times\) \(8\over 9\) = \(8\over 243\) Similar Questions The probability of India winning a test match against the west indies is 1/2 assuming independence …

A pair of fair dice is thrown independently three times. The probability of getting a score of exactly 9 twice is Read More »

Solution : A = {4. 5. 6}  and  B = {1, 2, 3, 4} \(A \cap B\) = 4 \(\therefore\)  \(P(A \cup B)\) = P(A) + P(B) – \(A \cap B\) \(\implies\) \(P(A \cup B)\) = 3/6 + 4/6 – 1/6 = 1 Similar Questions The probability of India winning a test match against the …

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 Read More »

Solution : We know that, P(A/B) = \(P(A \cap B)\over P(B)\)    …….(i) and P(B/A) = \(P(B \cap A)\over P(A)\)    ……….(ii) \(\therefore\)  P(B) = \(P(B/A).P(A)\over P(A/B)\) = \(1\over 3\) 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 …

It is given that the events A and B are such that P(A) = 1/4, P(A/B) = 1/2 and P(B/A) = 2/3. Then, P(B) is equal to Read More »

Solution : S = { 00, 01, 02, ……, 49 } Let A be the event that sum of the digits on the selected ticket is 8, then A = { 08, 17, 26, 35, 44 } Let B be the event that the product of the digits is zero. B = { 00, 01, …

One ticket is selected at random from 50 tickets numbered 00, 01, 02, ……, 49. Then, the probability that the sum of the digits on the selected ticket is 8, given that the product of these digits is zero equal to Read More »

Solution : Total number of cases = \(^9C_3\) = 84 Number of favourable cases = \(^3C_1\).\(^4C_1\).\(^2C_1\) = 24 \(\therefore\)  P = \(24\over 84\) = \(2\over 7\) 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 …

An urn contains nine balls of which three are red, four are blue and two are green. Three balls are drawn at random without replacement from the urn. The probability that the three balls have different colors is Read More »

Solution : \(P({A’ \cap B’\over C})\) = \(P(A’ \cap B’ \cap C)\over P(C)\) = \(P(C) – P(A \cap C) – P(B \cap C) + P(A \cap B\cap C)\over P(C)\)   ……..(i) Given, \(P(A \cap B\cap C)\)  = 0 and A, B and C are pairwise independent. \(\therefore\)  \(P(A \cap C)\) = P(A).P(C) and \(P(B \cap C)\) …

Let A, B and C are pairwise independent events with P(C) > 0 and \(P(A \cap B\cap C)\) = 0. Then \(P(A’ \cap B’/C)\) is equal to Read More »

Solution : Probability of guessing a correct answer, p = \(1\over 3\) and probability of guessing a wrong answer, q  = \(2\over 3\) So, the probability of guessing 4 or more correct answers is = \(^5C_4\) \(({1\over 3})^4\). \(2\over 3\) + \(^5C_5\) \(({1\over 3})^5\) = \(5.2\over {3^5}\) + \(1\over {3^5}\) = \(11\over {3^5}\) Similar Questions …

A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more correct answers just by guessing is Read More »

Question : If C and D are two events such that C \(\subset\) D and P(D) \(\ne\) 0, then the correct statement among the following is (a) P(C/D) \(\ge\) P(C) (b) P(C/D) < P(C) (c) P(C/D) = \(P(D)\over P(C)\) (d) P(C/D) = P(C) Solution : As P(C/D) = \(P(C \cap D)\over P(D)\) = \(P(C)\over P(D)\)  …

If C and D are two events such that C \(\subset\) D and P(D) \(\ne\) 0, then the correct statement among the following is Read More »