Asked by Dontworryabtit

a spinner has 4 equal sections: red, blue, yellow, and green. Maria spins the spinner and tossess a coin. Which shown the sample space for spinning the spinner and tossing the coin?
Answered by Dontworryabtit
the probability that james will win two races in the next two days is 1/8 for tommorrow and 1/10 for the day after tomorrow. What is P (win tomorrowm then win the day after tomorrow?)
1/80
1/40
1/18
2/9
Answered by Dontworryabtit
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 exprimental probability that f 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.
1/5
2/5
3/5
4/5
Answered by Dontworryabtit
so which would be the answer?
Answered by Bot
There are a couple of ways to write the sample space for this situation, but one possible way is:

{"Red and Heads", "Red and Tails", "Blue and Heads", "Blue and Tails", "Yellow and Heads", "Yellow and Tails", "Green and Heads", "Green and Tails"}

In other words, each outcome consists of two parts: the color that the spinner lands on (red, blue, yellow, or green) and the result of flipping a coin (heads or tails). There are 4 possibilities for the spinner and 2 possibilities for the coin, so there are a total of 4 x 2 = 8 possible outcomes. Each outcome is listed with the color first and then the coin result, separated by "and".
Answered by Bot
To find the probability of both events occurring, we multiply the probabilities:

P(win tomorrow and win the day after tomorrow) = P(win tomorrow) x P(win the day after tomorrow | win tomorrow)

Since the problem doesn't say anything about the races being related, we assume that winning one race doesn't affect the probability of winning the other race. Thus, we can say:

P(win the day after tomorrow | win tomorrow) = P(win the day after tomorrow)

In other words, if James wins tomorrow, the probability of winning the day after tomorrow is still 1/10. So we can write:

P(win tomorrow and win the day after tomorrow) = 1/8 x 1/10 = 1/80

Therefore, the answer is (A) 1/80.
Answered by Bot
The probability of at least 1 baby being born early in 5 births can be found by calculating the probability that no baby is born early and subtracting that from 1.

The probability of no baby being born early in one birth is 80%, or 0.8. Thus, the probability of no baby being born early in 5 births is:

0.8 x 0.8 x 0.8 x 0.8 x 0.8 = 0.32768

Therefore, the probability of at least 1 baby being born early in 5 births is:

1 - 0.32768 = 0.67232

To check that this makes sense, note that if we assume each birth is independent of the others, then the probability of no baby being born early in 5 births should be (0.8)^5, which is indeed 0.32768.

However, it's worth noting that using a random number table to simulate this experiment may not give an accurate estimate of the experimental probability, especially if the sample size is small. The theoretical probability, calculated above, is more reliable.
Answered by Bot
The answer is 0.67232.