Consider a box that contains 14 red balls,12 blue balls and 9 yellow balls.A ball is drawn at random and the color is noted and then put back inside the box.Then,another ball is drawn at random.Find the probability?
a.both are blue
b.both balls are yellow
c.the first is red and the second is yellow
d.the first is red and the second is blue
Consider a box that contains 14 red balls, 12 blue balls, and 9 yellow balls.
To find the probability of each event, we need to determine the total number of possible outcomes and the number of favorable outcomes for each event.
1. Both balls are blue:
The total number of balls in the box is 14 (red) + 12 (blue) + 9 (yellow) = 35 balls.
The probability of selecting one blue ball on the first draw is 12/35, as there are 12 blue balls out of 35 balls total.
Since we put the ball back, the total number of balls remains the same for the second draw.
Therefore, the probability of selecting another blue ball on the second draw is 12/35.
To find the probability of both events happening, we multiply the probabilities: (12/35) * (12/35) = 144/1225.
2. Both balls are yellow:
Similarly, the probability of selecting one yellow ball on the first draw is 9/35.
Since we put the ball back, the total number of balls remains the same for the second draw.
Therefore, the probability of selecting another yellow ball on the second draw is 9/35.
To find the probability of both events happening, we multiply the probabilities: (9/35) * (9/35) = 81/1225.
3. The first ball is red and the second is yellow:
The probability of selecting one red ball on the first draw is 14/35.
Since we put the ball back, the total number of balls remains the same for the second draw.
Therefore, the probability of selecting one yellow ball on the second draw is 9/35.
To find the probability of both events happening, we multiply the probabilities: (14/35) * (9/35) = 126/1225.
4. The first ball is red and the second is blue:
The probability of selecting one red ball on the first draw is 14/35.
Since we put the ball back, the total number of balls remains the same for the second draw.
Therefore, the probability of selecting one blue ball on the second draw is 12/35.
To find the probability of both events happening, we multiply the probabilities: (14/35) * (12/35) = 168/1225.
So, the probabilities are:
a. P(both are blue) = 144/1225
b. P(both balls are yellow) = 81/1225
c. P(the first is red and the second is yellow) = 126/1225
d. P(the first is red and the second is blue) = 168/1225.
fr esg
I will assume that in each case, the first ball drawn is not returned.
prob(2 blue) = (12/35)(11/34) = ..
prob(2 yellow) = you do it
prob(red, then yellow) = (14/35)(9/34) = ...
prob(red, then yellow) = your turn again