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 events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.

6/15 * 4/15 * 5/15 = ?

Without replacement, there is one less ball total in the following draws.

6/15 * 4/14 * 5/13 = ?

one bag contains 4 white balls,2 black balls.another bag contains 3 white balls 5 black balls.if one ball is drawn from each bag. find that

a) both are white?
b) both are black?
c)one is white and one is black?

The sample point for 1st bag and 2nd bag are given by n(S1)=4+2=6, n(S2)=3+5=8, If one ball is drawn from each bag, the probability that both are white is given by: P(w1 n w2) ie, probability of 1st being white and 2nd being white. P(w1 n w2)=4/6*3/8=1/4

P(both are Black),P(B1 n B2)=2/6*5/8=5/24
P(one is W and one is B)=P(W1 n B2) or P(B1 n W2)=(4/6*5/8)+(2/6*3/8)=13/24. NB: "n" is used as interception which means multiplying and "or" is the union which means addition.

To find the probability of drawing three balls in order (red, white, blue), we need to consider the number of favorable outcomes (balls drawn in the desired order) and the total number of possible outcomes.

(a) When each ball is replaced after drawing:
In this case, each time we draw a ball, put it back in the box, and mix the balls. So the total number of balls remains the same for each draw.

Total number of possible outcomes:
Since there are 6 red balls, 4 white balls, and 5 blue balls in the box, the total number of possible outcomes for each draw is 6 + 4 + 5 = 15.

Number of favorable outcomes:
To draw a red ball first, we have 6 options.
To draw a white ball second, we have 4 options left.
To draw a blue ball third, we have 5 options left.

Therefore, the number of favorable outcomes is 6 * 4 * 5 = 120.

Probability:
Now, we can calculate the probability by dividing the number of favorable outcomes by the total number of possible outcomes:
P(red, white, blue) = Number of favorable outcomes / Total number of possible outcomes
P(red, white, blue) = 120 / 15^3
P(red, white, blue) = 120 / 3375
P(red, white, blue) ≈ 0.0356 (approximately)

(b) When balls are not replaced after drawing:
In this case, after drawing each ball, we do not put it back into the box. So the total number of balls reduces with each draw.

Total number of possible outcomes:
For the first draw, there are 6 + 4 + 5 = 15 balls in total.
For the second draw, there are 14 balls left in total (as one ball has already been drawn).
For the third draw, there are 13 balls left (as two balls have already been drawn).

So the total number of possible outcomes for each draw is 15 * 14 * 13 = 2730.

Number of favorable outcomes:
Similar to the previous case, to draw a red ball first, we have 6 options.
To draw a white ball second, we have 4 options left.
To draw a blue ball third, we have 5 options left.

Therefore, the number of favorable outcomes is 6 * 4 * 5 = 120.

Probability:
Just like before, we can calculate the probability by dividing the number of favorable outcomes by the total number of possible outcomes:
P(red, white, blue) = Number of favorable outcomes / Total number of possible outcomes
P(red, white, blue) = 120 / 2730
P(red, white, blue) ≈ 0.0439 (approximately)

So, the probability of drawing three balls in order (red, white, blue) is approximately 0.0356 when each ball is replaced, and approximately 0.0439 when balls are not replaced.