A chocolate gift box contains 15 chocolates Six are Five Stars four are Fruits n Nut five are Dairy Milk After I have eaten the first chocolate a Fruit n Nut I pick another one The probability that I pick a Fruit n Nut again is

\frac{4}{15}

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\frac{1}{3}

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\frac{1}{5}

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\frac{3}{14}

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Solution

The correct option is D $\frac{3}{14}$
Total number of chocolates remaining after I pick and eat one 'Fruit 'n' Nut' chocolate = 15 - 1 = 14
Total no. of 'Fruit 'n' Nut' remaining = 4 -1 = 3
$∴$ Required Probability = $\frac{3}{14}$

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Picking a chocolate from a bag having 100 same identical dairy milk chocolates is a random experiment.

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A chocolate gift box contains 15 chocolates Six are Five Stars four are Fruits n Nut five are Dairy Milk After I have eaten the first chocolate a Fruit n Nut I pick another one The probability that I pick a Fruit n Nut again is

The correct option is D 314Total number of chocolates remaining after I pick and eat one 'Fruit 'n' Nut' chocolate = 15 - 1 = 14Total no. of 'Fruit 'n' Nut' remaining = 4 -1 = 3∴ Required Probability = 314

The correct option is D $\frac{3}{14}$
Total number of chocolates remaining after I pick and eat one 'Fruit 'n' Nut' chocolate = 15 - 1 = 14
Total no. of 'Fruit 'n' Nut' remaining = 4 -1 = 3
$∴$ Required Probability = $\frac{3}{14}$