Question

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.’

Solution

S = {(H, H), (H, T), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)} where (H, H) denotes that both the tosses result into head and (T, i) denote the first toss result into a tail and the number i appeared on the die for i = 1, 2, 3, 4, 5, 6.

therefore

the probabilities assigned to the 8 elementary events (H, H), (H, T), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6) are

1 fourth comma space 1 fourth comma space 1 over 12 comma space 1 over 12 comma space 1 over 12 comma space 1 over 12 comma space 1 over 12 comma space 1 over 12 space 1 over 12

respectively.

Let F be the event that ‘there is at least one tail’ and E be the event ‘the die shows a number greater than 4’. Then

straight F space equals space open curly brackets left parenthesis straight H comma space straight T right parenthesis comma space left parenthesis straight T comma space 1 right parenthesis comma space left parenthesis straight T comma space 2 right parenthesis comma space left parenthesis straight T comma space 3 right parenthesis comma space left parenthesis straight T comma space 4 right parenthesis comma space left parenthesis straight T comma space 5 right parenthesis comma space left parenthesis straight T comma space 6 right parenthesis close curly brackets straight E space equals space left curly bracket left parenthesis straight T comma space 5 right parenthesis comma space left parenthesis straight T comma space 6 right parenthesis right curly bracket space and space straight E intersection straight F space equals space left curly bracket left parenthesis straight T comma space 5 right parenthesis comma space left parenthesis straight T comma space 6 right parenthesis right curly bracket

Now,

straight P left parenthesis straight F right parenthesis space equals space straight P thin space open parentheses open curly brackets left parenthesis straight H comma space straight T right parenthesis close curly brackets close parentheses space plus space straight P space open parentheses open curly brackets left parenthesis straight T comma space 1 right parenthesis close curly brackets close parentheses space plus space straight P open parentheses open curly brackets left parenthesis straight T comma space 2 right parenthesis close curly brackets close parentheses

plus straight P open parentheses open curly brackets left parenthesis straight T comma space 3 right parenthesis close curly brackets close parentheses space plus space straight P space open parentheses open curly brackets left parenthesis straight T comma space 4 right parenthesis close curly brackets close parentheses space plus space straight P open parentheses open curly brackets left parenthesis straight T comma space 5 right parenthesis close curly brackets close parentheses space plus space straight P open parentheses open curly brackets left parenthesis straight T comma space 6 right parenthesis close curly brackets close parentheses

equals space 1 fourth plus 1 over 12 plus 1 over 12 plus 1 over 12 plus 1 over 12 plus 1 over 12 plus 1 over 12 equals space 1 fourth plus 6 over 12 space equals fraction numerator 3 plus 6 over denominator 12 end fraction space equals space 9 over 12 space equals 3 over 4

and

straight P left parenthesis straight E intersection straight F right parenthesis space equals space straight P open parentheses open curly brackets left parenthesis straight T comma space 5 right parenthesis close curly brackets close parentheses space plus space straight P open parentheses open curly brackets left parenthesis straight T comma space 6 right parenthesis close curly brackets close parentheses space equals space 1 over 12 plus 1 over 12 space equals 2 over 12 space equals 1 over 6

therefore space space space space straight P left parenthesis straight E space left enclose straight F right parenthesis space equals 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 fraction numerator begin display style 1 over 6 end style over denominator begin display style 3 over 4 end style end fraction space equals 1 over 6 cross times 4 over 3 space equals space 2 over 9

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Probability

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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