A Goldsmith can choose between three metals A, B and C. The probability that he chooses C is .

\frac{1}{3}

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\frac{2}{5}

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\frac{1}{4}

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Solution

The correct option is A $\frac{1}{3}$
There are 3 metals A, B and C.
So the total number of events = 3
Number of favourable events = 1
$Probability = \frac{N u m b e r o f f a v o u r a b l e e v e n t s}{T o t a l n u m b e r o f e v e n t s}$
$∴ Probability = \frac{1}{3}$

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A Goldsmith can choose between three metals A, B and C. The probability that he chooses C is .

The correct option is A 13There are 3 metals A, B and C. So the total number of events = 3 Number of favourable events = 1 Probability=Number of favourable eventsTotal number of events ∴Probability=13

The correct option is A $\frac{1}{3}$
There are 3 metals A, B and C.
So the total number of events = 3
Number of favourable events = 1
$Probability = \frac{N u m b e r o f f a v o u r a b l e e v e n t s}{T o t a l n u m b e r o f e v e n t s}$
$∴ Probability = \frac{1}{3}$