There are four marbles in a bag with the colors red, white, blue, and green. John pulls out one marble and tosses a coin. How many elements are there in the sample space?
A. 2
B. 4
C. 8
D. 16
There are 4 possible outcomes when John pulls out a marble. Then, for each of those outcomes, there are 2 possible outcomes when he tosses a coin. So the total number of elements in the sample space is 4 $\times$ 2 = 8.
Therefore, the answer is $\boxed{\textbf{(C)}\ 8}$.
Alli's ice cream shop offers 5 flavors and 10 toppings. Jali's ice cream shop offers 7 flavors and 7 toppings. Fernando's ice cream shop offers 9 flavors and 6 toppings. Laura's ice cream shop offers 13 flavors and 4 toppings.
If you want one flavor of ice cream and one topping, which shop gives you the most options?
A. Alli's ice cream shop
B. Jali's ice cream shop
C. Fernando's ice cream shop
D. Laura's ice cream shop
To find out which shop gives you the most options, simply add up the number of flavors and toppings at each shop.
- Alli's: 5 + 10 = 15 options
- Jali's: 7 + 7 = 14 options
- Fernando's: 9 + 6 = 15 options
- Laura's: 13 + 4 = 17 options
Therefore, Laura's ice cream shop gives you the most options if you want one flavor of ice cream and one topping. The answer is $\boxed{\textbf{(D)}\ \text{Laura's ice cream shop}}$.
The probability that James will win two races in the next two days is one-eighth for tomorrow and one-tenth for the day after tomorrow. What is P(win tomorrow, then win the day after tomorrow)?
A. start fraction 1 over 80 end fraction
B. start fraction 1 over 40 end fraction
C. Start Fraction 1 over 18 End Fraction
D. Start Fraction 2 over 9 End Fraction
Using the multiplication rule of probability, we can multiply the probability of winning tomorrow by the probability of winning the day after tomorrow to find the probability of winning both races:
\begin{align*}
P(\text{win tomorrow, then win the day after tomorrow}) &= P(\text{win tomorrow}) \cdot P(\text{win the day after tomorrow}) \\
&= \frac{1}{8} \cdot \frac{1}{10} \\
&= \frac{1}{80}.
\end{align*}
Therefore, the answer is $\boxed{\textbf{(A)}\ \frac{1}{80}}$.
Suppose 20% of babies born are born early, 50% are born on time, and 30% are born late. A nurse uses a random-number table to find the experimental probability that of 5 births, at least 1 baby will be born early. The digits 0 and 1 represent babies born early. The digits 2, 3, 4, 5, and 6 represent babies born on time. The digits 7, 8, and 9 represent babies born late.
23059 78234 01785 12359 26789
14568 24579 13579 01239 24589
03489 12456 01458 23567 01238
01235 34567 23478 13546 23589
Find the experimental probability that of 5 babies born, at least 1 will be born early.
A. one-fifth
B. two-fifths
C. three-fifths
D. start fraction 4 over 5 end fraction
We start by finding the probability that none of the 5 babies will be born early, using the fact that the probability of any one baby being born on time or late is 0.5 + 0.3 = 0.8:
\begin{align*}
P(\text{none born early}) &= 0.8 \cdot 0.8 \cdot 0.8 \cdot 0.8 \cdot 0.8 \\
&= 0.32768.
\end{align*}
Then, we can find the probability that at least one baby will be born early by subtracting this probability from 1:
\begin{align*}
P(\text{at least one born early}) &= 1 - P(\text{none born early}) \\
&= 1 - 0.32768 \\
&= 0.67232.
\end{align*}
Therefore, the experimental probability that of 5 babies born, at least 1 will be born early is about $\boxed{\textbf{(D)}\ \frac{4}{5}}$.