A basket contains 5 green lollipops, 12 red lollipops, and 7 orange lollipops. When a lollipop is taken from the basket, it is not replaced. What is P(orange, then orange)?
A.42/576
B.42/552
C.7/24
D.14/48
Ithink it is C...?
prob = (7/24)(6/23)
= 42/552
= 7/92
looks like B, but they should have reduced the fraction to lowest terms.
they usually dont reduce it, so i say B.
thank you
so is the right one the simplified version or no?
The simplified version is technically the correct answer, so the correct answer is B (42/552), but it's common for math problems to be left in unsimplified form as well.
To calculate the probability of drawing two orange lollipops in a row without replacement, we need to find the probability of drawing the first orange lollipop and then, given that, the probability of drawing the second orange lollipop.
First, we need to find the probability of drawing the first orange lollipop. There are a total of 24 lollipops in the basket (5 green + 12 red + 7 orange) and 7 of them are orange. Therefore, the probability of drawing the first orange lollipop is 7/24.
Next, we need to find the probability of drawing the second orange lollipop, given that the first one was already drawn and not replaced. After the first orange lollipop is drawn, there are now 23 lollipops left in the basket, and 6 of them are orange. Therefore, the probability of drawing the second orange lollipop is 6/23.
To find the probability of both events happening (drawing the first orange lollipop and then drawing the second orange lollipop), we multiply the probabilities together: (7/24) * (6/23) = 42/552.
So, the correct answer is B. 42/552.