Bunuel wrote:
In a contest, a bowl contains 10 keys, one of which will open a treasure chest and nine of which will not. If a contestant selects the key that opens the treasure chest, she wins the contents of that chest. If Anna is allowed to draw two keys, simultaneously and at random, from the bowl as the first contestant, what is the probability that she wins the prize?
A. 1/10
B. 1/9
C. 1/5
D. 19/90
E. 2/9
Kudos for a correct solution.
There are two ways in which the right key is taken out
(1st key is the right one & second wrong one) or (1st key is the wrong one & 2nd one is right)
Since it is "or" hence we will add the probabilities.
(
\(\frac{1}{10}\) *
\(\frac{9}{9}\)) + (
\(\frac{9}{10}\) *
\(\frac{1}{9}\))
\(\frac{1}{10}\) = Only one key is the right key. Total options are 10
\(\frac{9}{9}\) = After first key is taken out, there r 9 favorable options (all keys r wrong). Total options are also 9
\(\frac{9}{10}\) = For first key to be the wrong one, favorable options are nine. Total options are 10.
\(\frac{1}{9}\) = For second key to be correct one, only one favorable option is there out of available 9 options.
\(\frac{9}{90}\) + \(\frac{9}{90}\)
\(\frac{18}{90}\) =
\(\frac{1}{5}\)C is the answer.