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?
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
5/10=1/2
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
To find the probability that the first coin is a dime and the second coin is a nickel, we need to determine the total number of coins and the number of desirable outcomes.
Total number of coins in Gary's pocket = 10 (as given)
Number of desirable outcomes:
To calculate the probability, we need to calculate the probability of the first coin being a dime and the second coin being a nickel separately.
First coin being a dime:
Gary has a total of 5 dimes in his pocket, and since he does not look at the coin before placing it on the table, the probability of selecting a dime as the first coin is 5/10 or 1/2.
Second coin being a nickel:
After removing the first coin, Gary is left with 9 coins in his pocket. Out of these, he has 3 nickels. Therefore, the probability of selecting a nickel as the second coin, considering that the first coin was a dime, is 3/9.
To find the combined probability, we multiply the individual probabilities:
Probability = (Probability of the first coin being a dime) * (Probability of the second coin being a nickel)
= (1/2) * (3/9)
= 1/6
Therefore, the probability that the first coin Gary pulls is a dime and the second coin is a nickel is 1/6.