A quarter, nickel, and dime were tossed. Above are the possible outcomes. What is the possibility of getting at least two heads?
1/2
7/8
3/8
5/8
"Above are the possible outcomes" . No outcomes are listed in your post, but there
should be 8 of them. Count how many have 2 H's or 3 H's
then divide by 8
To determine the possibility of getting at least two heads when tossing a quarter, nickel, and dime, we need to count the number of outcomes that fulfill this condition and divide it by the total number of possible outcomes.
Let's first list all the possible outcomes when tossing the three coins:
1. Quarter - Heads, Nickel - Heads, Dime - Heads
2. Quarter - Heads, Nickel - Heads, Dime - Tails
3. Quarter - Heads, Nickel - Tails, Dime - Heads
4. Quarter - Heads, Nickel - Tails, Dime - Tails
5. Quarter - Tails, Nickel - Heads, Dime - Heads
6. Quarter - Tails, Nickel - Heads, Dime - Tails
7. Quarter - Tails, Nickel - Tails, Dime - Heads
8. Quarter - Tails, Nickel - Tails, Dime - Tails
Out of these eight possibilities, we can see that there are three outcomes (1, 3, and 5) that have at least two heads.
Therefore, the possibility of getting at least two heads is 3 out of 8.
To simplify this fraction, we can divide both the numerator and denominator by their greatest common divisor, which in this case is 1.
So, the possibility of getting at least two heads is:
3 / 8