Question 972967: Eleven contestants will try to win a car.
Nine women and two men are seated in eleven chairs. From left to right, a male occupies seat #9 and #11. A key is drawn from a basket containing 11 keys, one by one in the order they sit from left to right.
1. Find the probability the woman in seat #6 selects the correct key;
a.) With replacement of the keys.
b.) Without replacement of the keys.
2. Find the probability the man in seat #9 selects the correct key.
a.) With replacement of the keys.
b.) Without replacement of the keys
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! With replacement, I will assume that nobody knows until the end who won. Her chance is then 1 in 11. If the key has already been chosen, then her probability is zero, if that fact is made clear at the onset.
Without replacement, the first person has a (1/11) chance. The woman is sixth, and the five people before her cannot win. (10/11)*(9/10)*(8/9)*(7/8)*(6/7). That product is 30240 /55440=0.545
Her probability is 0.545.
Man in seat 9 has the same probability with replacement with the same concerns raised above. It is 1/11.
Continuing with the probabilities, (5/6)*(4/5)(3/4)=60/120. His probability is (0.545)*(0.5)=0.278.
Yes, you are correct. I got sloppy and thought 5/9 was the fractional equivalent of 0.545. It is not, and your answer is 3/11. Good pick up.
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