Consider the experiment of throwing a die, if a multiple of 3 co

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 Multiple Choice QuestionsShort Answer Type

771. Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that at least one is a girl?
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772.

A family has two children. What is the probability that both the children are boys given that at least one of them is a boy?

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773. Mother, father and son line up at random for a family picture
E : son on one end,    F : father in middle,  Determine P(E | F).
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774.

A couple has two children,
Find the probability that both children are males, if it is known that at least one of the children is male. 

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775.

A couple has two children,
Find the probability that both children are females, if it is known that the elder child is a female.

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 Multiple Choice QuestionsLong Answer Type

776.

An electronic assembly consists of two subsystems, say A and B. From the 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:
(i) P( A fails left enclose straight B space has space failed end enclose) (ii) P(A fails alone)

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777. Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event ‘the coin shows a tail’, given that ‘at least one die shows a 3’.


S = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6), (1, H), (1, T), (2, H), (2, T), (4, H), (4, T), (5, H), (5, T)}
The outcomes of S are not equally likely. First 12 outcomes are equally likely and are such that the sum of their probabilities is 2 over 6 space equals 1 third. So each of the first 12 outcomes has a probability equal to 1 third cross times 1 over 12 space equals space 1 over 36. Remaining eight outcomes are equally likely and are such that the sum of their probabilities is 4 over 6 space equals space 2 over 3.
So, each of these has a probability equal to fraction numerator 2 over denominator 3 cross times 8 end fraction equals 1 over 12.
Let  E: 'the coin shows a tail'
and F: 'at least one die shows up a 3',
therefore space space space space space straight E space equals space left curly bracket left parenthesis 1 comma space straight T right parenthesis comma space left parenthesis 2 comma space straight T right parenthesis comma space left parenthesis 4 comma space straight T right parenthesis comma space left parenthesis 5 comma space straight T right parenthesis right curly bracket
and  straight F space equals space left curly bracket left parenthesis 3 comma space 1 right parenthesis comma space left parenthesis 3 comma space 2 right parenthesis comma space left parenthesis 3 comma space 3 right parenthesis comma space left parenthesis 3 comma space 4 right parenthesis comma space left parenthesis 3 comma space 5 right parenthesis comma space left parenthesis 3 comma space 6 right parenthesis comma space left parenthesis 6 comma space 3 right parenthesis right curly bracket
therefore space space space space space space straight E intersection straight F space equals straight ϕ
therefore space space space space straight P left parenthesis straight E intersection straight F right parenthesis space equals space 0
Required probability  = straight P left parenthesis straight E space left enclose space straight F right parenthesis end enclose space equals space fraction numerator straight P left parenthesis straight E intersection straight F right parenthesis over denominator straight P left parenthesis straight F right parenthesis end fraction equals space fraction numerator 0 over denominator straight P left parenthesis straight F right parenthesis end fraction equals space 0.

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778. Consider the experiment of tossing a coin. If the coin shows head, toss it again but if it shows tail, then throw a die. Find the conditional probability of the event that ‘the die shows a number greater than 4’ given that ‘there is at least one tail.’

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 Multiple Choice QuestionsMultiple Choice Questions

779.

If straight P left parenthesis straight A right parenthesis space equals space 1 half comma space space space straight P left parenthesis straight B right parenthesis space equals space 0 comma space space then space straight P left parenthesis straight A space left enclose straight B right parenthesis is

  • 0

  • 1 half
  • not defined

  • not defined

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780. If A and B are events such that P(A|B) = P(B|A), then
  • A ⊂ B but A ≠ B
  • A = B
  • A ∩ B = ϕ
  • A ∩ B = ϕ
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