A bag consists of 4 different colored marbles.
The probability of drawing each is as follows:
P (Blue)=. 4 P (Red)= .2 P (Purple)= .3 P (Black)= .1
You draw a marble out of the bag and then replace it.
What is the probability that you draw:
a. A red marble first and a blue marble second?
b. A black marble first and a purple marble second?
c. A red or blue marble?
d. A gray marble?
as long as you REPLACE
.2 * .4
.1 * .3
.2 + .4 http://www-math.bgsu.edu/~albert/m115/probability/add_probs.html
0 because no gray in the bag
note
probability of blue or red or purple or black better be 1.000000 :)
add them up to check
To calculate the probabilities, we'll use the formula P(A and B) = P(A) * P(B), where P(A) represents the probability of event A and P(B) represents the probability of event B.
a. To find the probability of drawing a red marble first and a blue marble second, we multiply the probabilities of these two events: P(Red first and Blue second) = P(Red) * P(Blue) = 0.2 * 0.4 = 0.08 or 8%.
b. Similarly, to find the probability of drawing a black marble first and a purple marble second, we multiply the probabilities of these two events: P(Black first and Purple second) = P(Black) * P(Purple) = 0.1 * 0.3 = 0.03 or 3%.
c. To find the probability of drawing a red or blue marble, we'll use the formula P(A or B) = P(A) + P(B) - P(A and B). In this case, we have only the events of drawing red or blue marbles happening separately, so P(Red or Blue) = P(Red) + P(Blue) = 0.2 + 0.4 = 0.6 or 60%.
d. However, since there are no gray marbles mentioned, the probability of drawing a gray marble would be 0 (zero), as it is not one of the options available in the bag.
To find the probability of drawing different marbles in the given scenarios, we'll use the concept of independent events. Since the marble is replaced after each draw, the probability of each color remains the same for each draw.
a. Probability of drawing a red marble first and a blue marble second:
To find the probability, we need to multiply the probability of drawing a red marble first (0.2) by the probability of drawing a blue marble second (0.4). Since these events are independent, the multiplication rule applies.
P(Red then Blue) = P(Red) * P(Blue) = 0.2 * 0.4 = 0.08
Therefore, the probability of drawing a red marble first and a blue marble second is 0.08.
b. Probability of drawing a black marble first and a purple marble second:
Similar to part a, we multiply the probability of drawing a black marble first (0.1) by the probability of drawing a purple marble second (0.3) to find the probability of these events occurring together.
P(Black then Purple) = P(Black) * P(Purple) = 0.1 * 0.3 = 0.03
Thus, the probability of drawing a black marble first and a purple marble second is 0.03.
c. Probability of drawing a red or blue marble:
To find this probability, we need to add the individual probabilities of drawing a red marble (0.2) and a blue marble (0.4) since they are mutually exclusive events (i.e., they cannot occur at the same time).
P(Red or Blue) = P(Red) + P(Blue) = 0.2 + 0.4 = 0.6
Therefore, the probability of drawing a red or blue marble is 0.6.
d. Probability of drawing a gray marble:
As per the given information, there is no mention of a gray marble in the bag. Thus, the probability of drawing a gray marble is 0.
Hence, the probability of drawing a gray marble is 0.