The total number of remaining bulbs after the first customer bought 4 is 16 (6 amaryllis + 7 daffodils + 4 lilies + 3 tulips - 4 bought = 16 remaining). Out of these 16, there are still 6 amaryllis bulbs left.
If the next customer selects 1 bulb at random, the probability of getting an amaryllis bulb is:
6 amaryllis / 16 total remaining bulbs = 0.375 or 37.5%
Therefore, the closest answer is 38%.
A garden store has the following miscellaneous flower bulbs in a basket
6 amaryllis
7 daffodils
4 lilies
3 tulips
a customer bought 4 bulbs from the basket one of each type of flower/ If the next customer selects 1 of them remaong bulbs at randome. which is closest to the probablity that customer will get an amaryllis bulb
30%
31%
38%
45%
6 answers
Eli has 7 black pens and 5 blue pens in his des drawer. He aslo has 3 yellow highlighters, 2 green highlighters and 5 pink highlighters in his pencil case. If he chose one pen and one highlighter without looking. What is the probablity that he will get a black pen and he will not get a green highlighter?
7/15
117/1000
1/3
1/12
7/15
117/1000
1/3
1/12
The probability of Eli getting a black pen is 7/12 (since he has 7 black pens out of a total of 12 pens).
The probability of Eli not getting a green highlighter is 10/12 (since he has 10 highlighters that are not green out of a total of 10 highlighters).
To get the probability of both events happening (getting a black pen and not getting a green highlighter), we need to multiply the probabilities:
P(black pen and not green highlighter) = P(black pen) x P(not green highlighter)
P(black pen and not green highlighter) = (7/12) x (10/12)
P(black pen and not green highlighter) = 70/144
P(black pen and not green highlighter) = 35/72
Therefore, the probability of Eli getting a black pen and not getting a green highlighter is 35/72 or approximately 0.4861.
None of the given answer choices matches this result, so none of them is correct.
The probability of Eli not getting a green highlighter is 10/12 (since he has 10 highlighters that are not green out of a total of 10 highlighters).
To get the probability of both events happening (getting a black pen and not getting a green highlighter), we need to multiply the probabilities:
P(black pen and not green highlighter) = P(black pen) x P(not green highlighter)
P(black pen and not green highlighter) = (7/12) x (10/12)
P(black pen and not green highlighter) = 70/144
P(black pen and not green highlighter) = 35/72
Therefore, the probability of Eli getting a black pen and not getting a green highlighter is 35/72 or approximately 0.4861.
None of the given answer choices matches this result, so none of them is correct.
The probability that a dessert sold at a certain cafe contains chocalate is 75%. The probability that a dessert contains both chocolate and nuts is 23%. Find the probability that a randomly chosen chocolate dessert contains nuts.
30.7%
32.6%
17.3%
31/7%
30.7%
32.6%
17.3%
31/7%
We can use the formula:
P(A and B) = P(A) + P(B) - P(A or B)
where A is the event "the dessert contains chocolate" and B is the event "the dessert contains nuts".
We are given that P(A) = 0.75 and P(A and B) = 0.23. We want to find P(B | A), the probability that a chocolate dessert contains nuts, which is the same as the conditional probability P(A and B)/P(A).
To find P(B), we can use the formula:
P(A or B) = P(A) + P(B) - P(A and B)
We know that P(A or B) = P(A) = 0.75, and we can solve for P(B):
P(B) = (P(A or B) - P(A) + P(A and B)) / (1 - P(A))
P(B) = (0.75 - 0.75 + 0.23) / (1 - 0.75)
P(B) = 0.23 / 0.25
P(B) = 0.92
Now we can find P(B | A):
P(B | A) = P(A and B) / P(A)
P(B | A) = 0.23 / 0.75
P(B | A) = 0.307 or 30.7%
Therefore, the probability that a randomly chosen chocolate dessert contains nuts is 30.7%. The closest answer choice is 31.7%, which may be a typo.
P(A and B) = P(A) + P(B) - P(A or B)
where A is the event "the dessert contains chocolate" and B is the event "the dessert contains nuts".
We are given that P(A) = 0.75 and P(A and B) = 0.23. We want to find P(B | A), the probability that a chocolate dessert contains nuts, which is the same as the conditional probability P(A and B)/P(A).
To find P(B), we can use the formula:
P(A or B) = P(A) + P(B) - P(A and B)
We know that P(A or B) = P(A) = 0.75, and we can solve for P(B):
P(B) = (P(A or B) - P(A) + P(A and B)) / (1 - P(A))
P(B) = (0.75 - 0.75 + 0.23) / (1 - 0.75)
P(B) = 0.23 / 0.25
P(B) = 0.92
Now we can find P(B | A):
P(B | A) = P(A and B) / P(A)
P(B | A) = 0.23 / 0.75
P(B | A) = 0.307 or 30.7%
Therefore, the probability that a randomly chosen chocolate dessert contains nuts is 30.7%. The closest answer choice is 31.7%, which may be a typo.
An airline has 81% of his flights depart on schedule. It has 69% of its flights depart and arrive on schedule. Find the probability that a flight that departs on schedule also arrives on schedule.
1.25
0.85
0.45
0.43
1.25
0.85
0.45
0.43