How many balls should be drawn from each of the colors?
First of all we have 10 balls of which 2 are red, 3 are black and 4 are blue. If we can think carefully, the question is given that, when 3 balls are drawn, the balls are of different colors. There are 3 colors. So definitely one ball should be drawn from each of the color.
How many black balls can be selected from 4 black balls?
Initially let’s find out the number of ways in which we can select the black balls. Note that at least 1 black ball must be included in each selection. or 2 black balls from 4 black balls. or 3 black balls from 4 black balls. or 4 black balls from 4 black balls. Now let’s find out the number of ways in which we can select the red balls.
How many red and white balls are there in the box?
There are 3 boxes, the first one containing 1 white, 2 red and 3 black balls; the second one containing 2 white, 3 red and 1 black ball and the third one containing 3 white, 1 red and 2 black balls. A box is chosen at random and from it two balls are drawn at random. One ball is red and the other, white.
What are the odds of drawing 2 red balls from 3?
Show that the odds are 7 to 3 against drawing 2 red balls from a bag containing 3 red and 2 white balls. In a bag there are 12 balls in which 5 are red.
How many blue and red balls are there in the box?
A box contains 5 blue and 4 red balls. Two balls are drawn successively from the box without replacement, and it is noted that the second one is red. What is the probability that the first is also red?
What is the probability of picking two different colored balls?
The probability of picking 2 red balls is 3/7 times 2/6 which is 1/7. Thus the probability of picking two same colored balls is 3/7 (The sum of the two). Thus the probability of picking different colored balls is 1-3/7 which is 4/7.
How many ways can 3 balls out of total 9 be selected?
There are total 9 bcalls in the box of which 2 are red, 3 black & 4 blue. Now 3 balls out of total 9 balls ,can be selected in total C (9, 3) = 9!/6! 3! = 84 ways. Now 1 red, 1 black and 1 blue balls can be selected respectively from 2 red, 3 black & 4 blue balls in- [C (2, 1) × C (3, 1) × C (4, 1)] = 2×3×4 = 24 no. of ways.