The probability of a student of pass in Maths, Physics and Chemis

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 Multiple Choice QuestionsMultiple Choice Questions

221.

The probability distribution of X is
X 0 1 2 3

P(X)

0.3 k 2k 2k

The value of k is

  • 0.7

  • 0.3

  • 1

  • 0.14


222.

Two events A and B will be independent if

  • PA'  B' = 1 - PA1 - PB

  • PA + PB = 1

  • PA = PB

  • A and B are mutually exclusive


223.

There are four letters and four addressed envelopes. The chance that all letters are not despatched in the right envelope is

  • 19/24

  • 21/23

  • 23/24

  • None of these


224.

The mean and variance ofa random variable X having a binomial distribution are 4 and 2 respectively, then P(X = 1) is

  • 132

  • 116

  • 18

  • 14


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225.

If the mean and standard deviation in probability binomial distribution are 3 and 3/2 respectively, then the probability distribution is

  • 34 + 1412

  • 14 + 3412

  • 14 + 349

  • 34 + 149


226.

A student appears in the test I, II, III. Student is declared pass if he passes any two of three tests. The probability of the student of pass test I, II, III are p, q and 12. If the probability of student's success is 12, then

  • p = 1, q = 0

  • p = 23, q = 12

  • p, q have infinitly many values

  • All of the above


227.

If A and B are the events for which PA = 13, PB = 14 and PA  B = 15, then PAB is equal to

  • 45

  • 35

  • 75

  • 15


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228.

The probability of a student of pass in Maths, Physics and Chemistry are m, p and c respectively. Probability of passing atleast in one subject is 75% and atleast in two subjects, is 50% and in exactly two subject is 40%, then which relation is correct ?

  • p + m + c = 1920

  • p + m + c = 2720

  • p + m + c = 110

  • None of these


B.

p + m + c = 2720

Let A, B and C respectively denote the events that the student passes in Maths, Physics and Chemistry.

Given that, P(A) = m, P(B)= p, P(C) = c and P(passing in atleast one)

               = PA  B  C = 0.75     1 - PA'  B'  C' = 0.75 1 - PA' . PB' . PC' = 0.75 A, B, C are independent events

 0.75 = 1 - 1 - m1 - p1 - c 0.25 = 1 - m1 - p1 - c 1 - m + p + c + pm + pc +cm - pcm = 0.25      ...iAlso, Ppassing exactly in two subjects = 0.4 PA  B  C  A  B  C  A  B  C = 0.4 PA  B  C + PA  B  C + PA  B  C = 0.4 PAPBPC + PAPBPC + PAPBPC = 0.4 pm - pmc + pc - pmc + mc - pmc = 0.4 pm + mc + pc - 3pmc = 0.4                                        ...iiAgain, P(passing atleast in two subjects) = 0.5  PA  B  C + PA  B  C + PA  B  C = 0.5 pm - pcm + pc - pcm + cm - pcm + pcm = 0.5 pm + pc + mc - 2pcm = 0.5                                       ...iiiOn solvmg Eqs. (i), (ii) and (iii), we getp + m + c = 1.35 = 2720


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229.

A dice is rolled 6 times, if the appearance of an even number on it is assumed a success, then the probability of getting 5 success is

  • 132

  • 332

  • 116

  • 532


230.

The probability that at least one ofthe events A and B occurs is 0.6. If A and B occur simultaneously with probability 0.2, then PA + PB will be

  • 1.5

  • 1.3

  • 1.2

  • 0.8


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