3 balls are drawn successively from a box containing 6red,4 white and 5 blue balls.Find the probability that they are drawn in order red,white and blue if:(a)each ball is replaced.(b)balls are not replaced
If the balls are not replaced you are dealing with dependent events, that is, the third event depends on what happened in the first, and the second depends on what happened first. So ...
prob(red, then white, then blue) = (6/15)(4/14)(5/13) = ...
If the balls are replaced, they are independent events. In this case the denominators all remain at 15
To find the probability of drawing the balls in order (red, white, blue), we need to consider the cases where the balls are replaced and not replaced separately.
(a) Each ball is replaced:
In this case, since each ball is replaced after it is drawn, the probability of drawing a particular ball does not change. Therefore, the probability of drawing the first ball as red is 6/15 (since there are 6 red balls out of a total of 15 balls), the probability of drawing the second ball as white is 4/15 (since there are 4 white balls remaining out of a total of 15 balls), and the probability of drawing the third ball as blue is 5/15 (since there are 5 blue balls remaining out of a total of 15 balls).
To find the probability of all three events occurring in order, we multiply the probabilities together:
P(red, white, blue) = (6/15) * (4/15) * (5/15) = 120/3375 = 8/225
So, the probability of drawing the balls in order (red, white, blue) with replacement is 8/225.
(b) Balls are not replaced:
In this case, after each ball is drawn, it is not replaced, so the probability of drawing each subsequent ball changes.
The probability of drawing the first ball as red is 6/15 (same as before). However, since the first ball is not replaced, the probability of drawing the second ball as white is 4/14 (since there are 4 white balls remaining out of a total of 14 balls). Similarly, the probability of drawing the third ball as blue is 5/13 (since there are 5 blue balls remaining out of a total of 13 balls).
Again, multiplying the probabilities together:
P(red, white, blue) = (6/15) * (4/14) * (5/13) = 120/2730 = 8/182
So, the probability of drawing the balls in order (red, white, blue) without replacement is 8/182.
To find the probability of drawing the balls in a specific order, we need to consider the total number of possible outcomes and the favorable outcomes.
(a) When each ball is replaced:
In this case, every time a ball is drawn, it is replaced back into the box before the next draw. The probability of drawing a red ball on any given draw is 6/15, as there are 6 red balls out of a total of 15 balls. Similarly, the probabilities of drawing a white ball and a blue ball are 4/15 and 5/15, respectively.
Since the balls are drawn successively, the probabilities are multiplied together:
P(red) * P(white) * P(blue) = (6/15) * (4/15) * (5/15)
(b) When balls are not replaced:
In this case, after each ball is drawn, it is not replaced back into the box before the next draw. The probabilities for the first draw remain the same as in part (a): P(red) = 6/15.
However, for the second draw, there are now only 14 balls left in the box (since one was already drawn). Therefore, the probability of drawing a white ball becomes 4/14.
Similarly, for the third draw, there are only 13 balls left in the box, so the probability of drawing a blue ball becomes 5/13.
Therefore, the probability of drawing the balls in order is:
P(red, white, blue) = (6/15) * (4/14) * (5/13)
Note: In both cases, we assume that the draws are independent events, meaning that the outcome of one draw does not affect the probabilities of the subsequent draws.