A basket contains 3 red balls 5 blue balls and 7 green balls. Two balls are picked one after the other without replacement find the probability that. (a) Both are red. (b) First is blue, the other is green. (c) One is blue, the other is green. (d) Both are of different colours. (e) They are of the same colours.
a) prob(RR) = (3/15)(2/14) = 1/35
b) Prob(B, then G) = (5/15)(7/14) = 1/6 <---- the order was important
c) Prob( B, G) order not important = P(B,G) + P(G,B) = ....
do e) first
e) prob( same colour) = Prob(R,R) + Prob(G,G) + prob(B,B)
= (3/15)(2/14) + (7/15)(6/14) + (5/15)(4/14) = ....
d) Prob(2 different colours) = 1 - (answer to e) )
I need full answer
I need the complete solving
I need the complete solving
Full solution plz
Well, well, well! Let's tackle these probability questions together, shall we?
(a) Both are red:
Well, if we pick one red ball first, there are 3 red balls left out of a total of 14 remaining balls. So the probability of picking another red ball is 3/14. Since we multiply probabilities when events are independent, the probability of picking both red balls is (3/15) * (3/14) = 3/70.
(b) First is blue, the other is green:
We have 5 blue balls and 7 green balls, so the probability of picking a blue ball first is 5/15. Once the first ball is removed, we are left with 14 balls, including 7 green ones. So the probability of picking a green ball after picking a blue ball is 7/14. Multiplying these probabilities together, we get (5/15) * (7/14) = 1/6.
(c) One is blue, the other is green:
This scenario can happen in two ways: either we pick a blue ball first and a green ball second, or the other way around. We've already calculated the probabilities for these individual events in the previous question. So, we just add them up: 1/6 + 1/6 = 1/3.
(d) Both are of different colors:
Here, we want to consider the cases where we pick a blue ball first and then either a red or a green ball, or where we pick a green ball first and then either a red or a blue ball next. The probability of picking a blue ball first, and either a red or green ball second is (5/15) * (10/14) = 1/3. The same calculation applies for picking a green ball first. Since these two cases are independent, we just add them up: 1/3 + 1/3 = 2/3.
(e) They are of the same color:
Here, we can either have both red balls, both blue balls, or both green balls. We've already calculated the probabilities for each of these cases in previous questions. So, we add them up: 3/70 + 1/15 + (7/15 * 6/14) = 1/7.
Well, I hope these answers brought a smile to your face!
To find the probabilities of the different scenarios, we need to calculate the total number of possible outcomes and the number of favorable outcomes for each scenario.
Total number of balls in the basket: 3 red + 5 blue + 7 green = 15 balls.
(a) Probability that both are red:
The probability of drawing a red ball on the first draw is 3/15 (since there are 3 red balls out of 15 total balls). After removing one red ball, there are 2 red balls out of 14 total remaining balls for the second draw. Therefore, the probability of drawing a red ball on the second draw is 2/14.
To find the probability of both events occurring, we multiply the probabilities: (3/15) * (2/14) = 6/210 = 1/35.
(b) Probability that the first is blue, and the other is green:
The probability of drawing a blue ball on the first draw is 5/15. After removing a blue ball, there are 7 green balls left out of 14 total remaining balls. Thus, the probability of drawing a green ball on the second draw is 7/14.
To find the probability of both events occurring, we multiply the probabilities: (5/15) * (7/14) = 35/210 = 1/6.
(c) Probability that one is blue, and the other is green:
There are two possible scenarios for this: either the first draw is blue and the second is green, or the first draw is green and the second is blue.
Probability of drawing a blue ball first and a green ball second: (5/15) * (7/14) = 1/6 (as calculated in part b).
Probability of drawing a green ball first and a blue ball second: (7/15) * (5/14) = 1/6.
Since these scenarios are mutually exclusive (they cannot both occur at the same time), we can add their probabilities: 1/6 + 1/6 = 2/6 = 1/3.
(d) Probability that both are of different colors:
To calculate this probability, we need to consider the possible combinations of colors. There are three options: (blue, red), (blue, green), and (red, green). Let's calculate the probability for each combination separately and sum them up.
Probability of drawing a blue ball first and a red ball second: (5/15) * (3/14) = 1/14.
Probability of drawing a blue ball first and a green ball second: (5/15) * (7/14) = 1/6 (as calculated in part b).
Probability of drawing a red ball first and a green ball second: (3/15) * (7/14) = 1/10.
Summing up these probabilities: 1/14 + 1/6 + 1/10 = 5/35 = 1/7.
(e) Probability that they are of the same color:
We can calculate this probability by subtracting the probability of both different colors (as calculated in part d) from 1 (since it covers all possible outcomes).
Probability of them being of the same color = 1 - 1/7 = 6/7.
Therefore,
(a) Probability that both are red = 1/35
(b) Probability that the first is blue, and the other is green = 1/6
(c) Probability that one is blue, and the other is green = 1/3
(d) Probability that both are of different colors = 1/7
(e) Probability that they are of the same color = 6/7