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Probability

Short Answer Type

761.

A coin is tossed three times, where
E : at least two heads , F : at most two heads

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762.

A coin is tossed three times, where
E : at most two tails , F : at least one tail

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763. A black and a red dice are rolled.Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.

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764. A black and a red dice are rolled. Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.

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Long Answer Type

765. Given that the two numbers appearing on throwing two dice are different. Find the probability of the event ‘the sum of numbers on the dice is 4’.

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766. An instructor has a question bank consisting of 300 easy True/False questions, 200 difficult True/False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple choice question?

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Short Answer Type

767. Find the probability of drawing a king, when a card is drawn from a well shuffled pack of cards. It is also given that the card drawn is a face card.

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768.

Two coins are tossed once, where
E : tail appears on one coin, F : one coin shows head
Determine P(E | F).

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769.

Two coins are tossed once, where
E : no tail appears, F : no head appears
Determine P(E | F).

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770. 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 the youngest is a girl?

Let the first child be denoted by capital letter and the second i.e. younger one by a smaller letter.
∴ S = {Bb, Bg, Gb, Gg}
Let E : both children are girls
∴ E = {Gg}
Let F : the youngest is a girl
$therefore space space space space straight F space equals open curly brackets Bg comma space Gg close curly brackets therefore space space space straight E intersection straight F space equals space open curly brackets Gg close curly brackets therefore space space space space straight P left parenthesis straight F right parenthesis space equals space 2 over 4 comma space space space straight P left parenthesis straight E intersection straight F right parenthesis space equals space 1 fourth$
Required proability = $straight P left parenthesis straight E space left enclose straight F right parenthesis space equals space fraction numerator straight P left parenthesis straight E intersection straight F right parenthesis over denominator straight P left parenthesis straight F right parenthesis end fraction space equals space fraction numerator begin display style 1 fourth end style over denominator begin display style 2 over 4 end style end fraction space equals 1 half$

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