At the beginning, he had 10 coins and drew one. For this one to be a dime, it has a probability of 5/10 (5 dimes out of 10 coins in total).
So p1=1/2
Then, after drawing that coin, he has 9 total coins in his pocket, then he drew one. For this coin to be a nickle, it has a probability of 3/9 (3 nickels out of 9 total coins).
So p2=1/3
Thus, p=p1*p2=(1/2)*(1/3)
Therefore, p=1/6
Gary has 10 coins in his pocket.
2 quarters
5 dimes
3 nickels
without looking, Gary pulls 1 coin from
his pocket and puts it on a table. Then,
he pulls one more coin from his pocket.
What is the probability that the first
coin is a dime and the second coin is a
nickle?
3 answers
5/10=1/2
3/9=1/3
1/2*1/3=1/6✌️
3/9=1/3
1/2*1/3=1/6✌️
Pro(dime) = 5/10
Without replacement total left= 9
Thus prob(nickel) = 3/9
Therefore
Pro(dime&nickel)=(5/10)(3/9)= 1/6
Without replacement total left= 9
Thus prob(nickel) = 3/9
Therefore
Pro(dime&nickel)=(5/10)(3/9)= 1/6
2 answers