A man has 3 pennies, 2 nickles, and 3 dimes. If he draws 1 coin randomly then replaces it and selects again randomly. What is the probability of getting less than 10 cents if his 1st draw was a penny?
So he already has 1 penny, which means he can only draw another penny or another nickel
Prob(as stated) = 3/8 + 2/8
= 5/8
Probability
To find the probability of getting less than 10 cents on the second draw, given that the first draw was a penny, we need to consider the possible outcomes and their probabilities.
First, let's calculate the probability of drawing a penny on the first draw. Since there are a total of 3 pennies out of 8 coins (3 pennies + 2 nickels + 3 dimes), the probability of drawing a penny on the first draw is 3/8.
Now, let's examine the possible outcomes on the second draw:
1. Drawing a penny again: There are still 3 pennies remaining, and the probability of drawing a penny again is 3/8.
2. Drawing a nickel: There are 2 nickels remaining, and the probability of drawing a nickel is 2/8.
3. Drawing a dime: There are 3 dimes remaining, and the probability of drawing a dime is 3/8.
Since we want to find the probability of getting less than 10 cents, we only need to consider the outcomes of drawing pennies and nickels. The probability of getting less than 10 cents will be the sum of the probabilities of drawing a penny again and drawing a nickel.
P(probability of getting less than 10 cents | first draw was a penny) = P(drawing a penny again) + P(drawing a nickel)
P(drawing a penny again) = 3/8
P(drawing a nickel) = 2/8
Adding these probabilities together:
P(probability of getting less than 10 cents | first draw was a penny) = 3/8 + 2/8 = 5/8
Therefore, the probability of getting less than 10 cents on the second draw, given that the first draw was a penny, is 5/8.