A box contains 5 white tags numbered 1 to 5, 5 red tags numbered from 1 to 5, and 5 green tags numbered from 1 to 5. If two tags are selected at random and without replacement what is the probability that both are the same number?

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A box contains 5 white tags numbered 1 to 5, 5 red tags numbered from 1 to 5, and 5 green tags numbered from 1 to 5. If two tags are selected at random and without replacement what is the probability that both are the same number?

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Answer 1 · 2016-01-23T02:03:38

$\frac{1}{7}$ $1/7$

Explanation:

The probability of an event occurring given a finite number of possible outcomes may be calculated as
$P (e v e n t) = \frac{number of ways event could occur}{number of possible results}$ $P(event) = "number of ways event could occur"/"number of possible results"$

There are $15$ $15$ possible ways to pick the same number, as you can have either white and red, red and green, or white and green for any one of the $5$ $5$ numbers.

The number of ways to pick two items from a set of fifteen is

$C (15, 2) = \frac{15!}{2! (15 - 2)!} = \frac{15 \cdot 14}{2} = 105$ $C(15, 2) = (15!)/(2!(15-2)!) = (15*14)/2 = 105$

Thus the probability of choosing the same number is

$\frac{15}{105} = \frac{1}{7}$ $15/105 = 1/7$

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A box contains 5 white tags numbered 1 to 5, 5 red tags numbered from 1 to 5, and 5 green tags numbered from 1 to 5. If two tags are selected at random and without replacement what is the probability that both are the same number?

1 Answer

Explanation:

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