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%
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%.
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
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 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%
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.
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
To find the probability that the next customer will select an amaryllis bulb, we need to calculate the ratio of the number of amaryllis bulbs to the total number of remaining bulbs.
Initially, there are 6 amaryllis bulbs in the basket. After the first customer bought one of each type of flower, there are 6-1 = 5 amaryllis bulbs remaining.
The total number of remaining bulbs is 6+7+4+3-4 = 16 (subtracting 4 since the first customer already bought 4 bulbs).
Therefore, the probability that the next customer will get an amaryllis bulb is 5/16 * 100% = 31.25%.
So, the closest option to the probability is 31%.