There are four red, three yellow and five blue marbles in a bag. What is the probability of selecting a blue marble?
There are a total of 4 + 3 + 5 = 12 marbles in the bag. Out of these, five a blue. Thus the number of favorable outcomes is 5 (since selecting any one of the five blue marbles gives the favorable outcome) and the total number of possible outcomes is 12 (since any one of the twelve marbles can be chosen). Applying the formula for theoretical probability,
`P("blue marble") = "Number of favorable outcomes"/"Total number of possible outcomes"`
`P("blue marble") = 5/12`
Thus the probability of selecting a marble that is blue is `5/12`.
From the same bag mentioned above, you pick one marble, replace it in the bag, and then pick another marble. What is the probability that both marbles will be red?
First find the probability of selecting a red marble from the bag. There are a total of 12 marbles, thus the total number of possible outcomes is 12. Further, there are four red marbles in the bag. Thus there are four favorable outcomes.
`P("red") = "Number of favorable outcomes"/"Total number of possible outcomes" = 4/12 = 1/3`
After selecting the first marble and then replacing it in the bag, you select another marble. Since the first marble was replaced, therefore the total number and number of red marbles is same. Thus the probability of selecting a red marble in the second attempt is the same as that in the first attempt, `1/3`.
Now find the combined probability of both marbles being red.
`P("both red") = P("first red and second red")`
The word 'and' in the above statement implies that since the two events of selecting the marbles are independent events, their combined probabilities is the product of their individual probabilities. Thus,
`P("both red") = 1/3 * 1/3 = 1/9`
Thus the probability of selecting two red marbles by replacing the first marble in the bag is `1/9`.
In the same question as above, the second marble is selected without replacing the first marble in the bag. Then find the probability of getting two red marbles.
This question is different from the above one in that it does not replace the first marble in the bag. Thus, the probability of getting a red marble in the second attempt changes.
The probability of the first marble selected being red is the same as before, `1/3`, since total number of marbles and number of red marbles is the same.
`P("first red") = 1/3`
After selecting the first marble, 11 marbles are left in the bag. Further, if we assume that the first marble selected was red, then 3 red marbles are left in the bag. The probability of getting a red marble from the bag now is
`P("second red") = "Number of red marbles left in the bag"/"Total number of marbles left in the bag" = 3/11`
Now the combined probability of the first marble being red and the second marble being red is the product of their individual probabilities since both are independent events.
`P("first red and second red") = P("first red") * P("second red")`
` = 1/3 * 3/11 = 1/11`
Thus, the probability of getting two red marbles from the bag without replacing the first one is `1/11`.
There are a total of 4 + 3 + 5 = 12 marbles in the bag. Out of these, five a blue. Thus the number of favorable outcomes is 5 (since selecting any one of the five blue marbles gives the favorable outcome) and the total number of possible outcomes is 12 (since any one of the twelve marbles can be chosen). Applying the formula for theoretical probability,
`P("blue marble") = "Number of favorable outcomes"/"Total number of possible outcomes"`
`P("blue marble") = 5/12`
Thus the probability of selecting a marble that is blue is `5/12`.
From the same bag mentioned above, you pick one marble, replace it in the bag, and then pick another marble. What is the probability that both marbles will be red?
First find the probability of selecting a red marble from the bag. There are a total of 12 marbles, thus the total number of possible outcomes is 12. Further, there are four red marbles in the bag. Thus there are four favorable outcomes.
`P("red") = "Number of favorable outcomes"/"Total number of possible outcomes" = 4/12 = 1/3`
After selecting the first marble and then replacing it in the bag, you select another marble. Since the first marble was replaced, therefore the total number and number of red marbles is same. Thus the probability of selecting a red marble in the second attempt is the same as that in the first attempt, `1/3`.
Now find the combined probability of both marbles being red.
`P("both red") = P("first red and second red")`
The word 'and' in the above statement implies that since the two events of selecting the marbles are independent events, their combined probabilities is the product of their individual probabilities. Thus,
`P("both red") = 1/3 * 1/3 = 1/9`
Thus the probability of selecting two red marbles by replacing the first marble in the bag is `1/9`.
In the same question as above, the second marble is selected without replacing the first marble in the bag. Then find the probability of getting two red marbles.
This question is different from the above one in that it does not replace the first marble in the bag. Thus, the probability of getting a red marble in the second attempt changes.
The probability of the first marble selected being red is the same as before, `1/3`, since total number of marbles and number of red marbles is the same.
`P("first red") = 1/3`
After selecting the first marble, 11 marbles are left in the bag. Further, if we assume that the first marble selected was red, then 3 red marbles are left in the bag. The probability of getting a red marble from the bag now is
`P("second red") = "Number of red marbles left in the bag"/"Total number of marbles left in the bag" = 3/11`
Now the combined probability of the first marble being red and the second marble being red is the product of their individual probabilities since both are independent events.
`P("first red and second red") = P("first red") * P("second red")`
` = 1/3 * 3/11 = 1/11`
Thus, the probability of getting two red marbles from the bag without replacing the first one is `1/11`.
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