Assuming the coins are randomly selected, the probability of pulling a dime first is the number of dimes (4) divided by the total number of coins (10).

p(dime first) = 4/10 = 2/5

Then, having drawn a dime, there are 9 coins left, of which 5 are nickels. The probability of randomly choosing a nickel is 5/9.

The joint probability of these two events occurring sequentially is the product of their probabilities:

p(dime then nickel) = (2/5)×(5/9) = 2/9

_____

<em>Alternate solution</em>

You can go at this another way. You can list all the pairs of coins that can be drawn. There are 90 of them: 10 first coins and, for each of those, 9 coins that can be chosen second. Of these 90 possibilities, there are 4 dimes that can be chosen first, and 5 nickels that can be chosen second, for a total of 20 possible dime-nickel choices out of the 90 total possible outcomes.

p(dime/nickel) = 20/90 = 2/9

8 0

3 years ago

There are 38 students. If there are 6 seats at each table, how many tables would be needed for all of the students to be able to

Leto [7]

Answer:

7 tables

Step-by-step explanation:

So if 6 students can fit at each table, you divide 38 by 6. You get 6 1/3 ,but since you can't get 1/3 of a table, you have to get 7

4 0

3 years ago