Assume a box contains three red jelly beans and two green ones. We consider the event that a red bean is drawn. Suppose I pick a jelly bean from the box without looking. I record the color but do not put the bean back in the box. Then I choose a bean again. What is the probability of getting two red beans?
Enter fraction in lowest terms.
P(red,red) = 3/5 * 2/4
You don't really need to record the color, since the bean was not replaced. Unless you have eaten it, you can just look at the two beans drawn.
What is the probability of second being red, if I ate the first one???
To solve this problem, we can use the concept of conditional probability.
Let's break down the problem step by step:
Step 1: Calculate the probability of drawing a red bean on the first pick.
Out of the total number of jelly beans in the box (3 red + 2 green = 5), the probability of drawing a red bean on the first pick is 3/5.
Step 2: Calculate the probability of drawing a red bean on the second pick, given that a red bean was already picked on the first pick.
After picking one red bean, there will be 2 red beans and 4 total beans left in the box. Therefore, the probability of drawing a red bean on the second pick, given that a red bean was already picked, is 2/4 or simplified to 1/2.
Step 3: Multiply the probabilities.
To calculate the probability of both events occurring (drawing two red beans in a row), we multiply the probabilities from step 1 and step 2:
(3/5) * (1/2) = 3/10
Therefore, the probability of drawing two red beans in a row is 3/10.
To calculate the probability of getting two red beans, we need to use the concept of conditional probability.
Let's break down the problem step by step:
Step 1: Calculate the probability of picking a red bean on the first draw.
There are a total of 5 jelly beans in the box, 3 of which are red. Therefore, the probability of picking a red bean on the first draw is 3/5.
Step 2: Calculate the probability of picking a red bean on the second draw given that the first bean was red.
After the first draw, we are left with 4 jelly beans in the box, 2 of which are red. Therefore, the probability of picking a red bean on the second draw, given that the first bean was red, is 2/4.
Step 3: Multiply the probabilities from both steps.
To find the probability of both events happening (picking a red bean on the first draw and then picking a red bean on the second draw), we multiply the probabilities together: (3/5) * (2/4) = 6/20.
Now, we should simplify the fraction 6/20 to its lowest terms. The greatest common divisor (GCD) of 6 and 20 is 2. Dividing both the numerator and the denominator by 2, we get 3/10.
Therefore, the probability of getting two red beans is 3/10.