4 red marbles, 3 green, 2 yellow. you pick 2 at random.What is probability of a. 1 yellow & 1 red
b. 2 green
is the first marble picked returned before the second is picked ?
- very important.
no
prob(yellow, then red)
= (2/9)(4/8)
= 1/9
follow the same kind of reasoning for your second part
To calculate the probability of events, we need to find the ratio of the favorable outcomes to the total number of possible outcomes.
a. Probability of picking 1 yellow and 1 red marble:
First, let's calculate the total number of possible outcomes. Since we are picking 2 marbles at random, there are (4+3+2) = 9 marbles in total. So, the total number of possible outcomes is C(9, 2) = 9! / (2! * (9-2)!) = 36.
Next, let's find the number of favorable outcomes, which is the number of ways we can choose 1 yellow marble and 1 red marble out of the 2 yellow and 4 red marbles. This can be calculated as C(2, 1) * C(4, 1) = 2 * 4 = 8.
Therefore, the probability of picking 1 yellow and 1 red marble is 8/36, which simplifies to 2/9.
b. Probability of picking 2 green marbles:
Similar to the previous calculation, let's calculate the total number of possible outcomes. Since there are 2 green marbles, the total number of possible outcomes is C(2, 2) = 1.
The number of favorable outcomes, which is the number of ways we can choose both green marbles out of the 3 green marbles available, is C(3, 2) = 3.
Therefore, the probability of picking 2 green marbles is 3/1, which simplifies to 3.
Note: The probabilities calculated in both cases are fractions and can be further simplified if necessary.