finding n of binomial distribution given the probability

 finding n of binomial distribution given the probability from miscellaneous 12th cbse ncert mathematics

How many times must a man toss a fair coin so that the probability of having at least one head is more than 90%?

Let n be the required number of times the man must toss a fair coin so that the probability of having at least one head is more than 90%

Let X be the number of heads obtained when the coin is tossed n times.

Assuming X follows binomial distribution with

n=n

p = (½)

q = 1 – p = (½)

P[X=r] = _nC_r p^rq^(n-r) , r = 0,1,2,...,n

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positive determinant from miscellaneous probability questions ncert cbse 12th mathematics

positive determinant and electronic assembly with two subsystems question from miscellaneous probability questions ncert cbse 12th mathematics

If each element of a second order determinant is either zero or one, what is the probability that the value of the determinant is positive?

There are 4 entries in the second order determinant ,each of which can be filled in two ways with either 0 or 1

Such determinants can be constructed in 2⁴ ways ( (2^4) ways)

Therefore if S is the sample space n(S) = (2^4) = 16

Let E be the event that the value of the selected determinant is positive. 

Required probability = P(E) = [n(E)] / [n(S)] = 3 /16

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An electronic assembly consists of two subsystems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known

P(A fails) = 0.2 P(B fails alone) = 0.15 P(A and B fail) = 0.15

Evaluate the following probabilities P(A fails|B has failed) ; P(A fails alone)

P(B fails) =P(A and B fail together ) + P(B fails alone) = 0.15+0.15 = 0.3

P(A fails|B has failed) = [P(A and B fail)] / [P(B fails ) ] = [0.15 / 0.30 =(1/2) = 0.5

P(A fails alone) = P(A fails) - P(A and B fail together ) = 0.2 – 0.15 = 0.05

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