there are 4 white balls and 6 red balls in a sack. If two balls are taken out successively ( the first ball is not replaced ) what is the probability that the balls are drawn are of different colors?
pr= prob second white given first red, plus prob second red given first white.
pr= 4/9*6/10 + 6/9*4/10= 48/90
A box contains 10 balls of which 6 are red (R) and 4 are white (W) . Three balls are taken from the box at random, one after the other. The probability that the 3 balls drawn are in the order RRW is equal to,
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Ah, the classic "battle of the balls" scenario! Let's calculate the probability of drawing balls of different colors.
To find the probability, we need to consider the total number of possible outcomes and the number of favorable outcomes.
First, let's determine the total number of possible outcomes. We draw two balls successively (without replacement) from a sack with 4 white balls and 6 red balls. So, the total number of balls available for the first draw is 10, and for the second draw, it's 9.
Therefore, the total number of possible outcomes is 10 * 9 = 90.
Now, let's determine the number of favorable outcomes, i.e., the number of ways we can draw balls of different colors.
First, we can choose one white ball from the 4 available, and then we can choose one red ball from the 6 available, or vice versa. So, the number of favorable outcomes is 4 * 6 + 6 * 4 = 48.
Finally, we can calculate the probability by dividing the number of favorable outcomes by the total number of possible outcomes:
Probability = Number of favorable outcomes / Total number of possible outcomes
= 48 / 90
≈ 0.5333 (rounded to 4 decimal places)
Therefore, the probability of drawing balls of different colors is approximately 0.5333 or 53.33%.
Remember, probabilities are like clown noses - they might seem funny, but they're serious business!
To find the probability of drawing two balls of different colors, we need to determine the total number of possible outcomes and the number of favorable outcomes.
First, let's calculate the total number of outcomes. In the first draw, we have 10 balls in the sack. So, there are 10 possible outcomes for the first draw.
For the second draw, since the first ball is not replaced, we have 9 balls remaining in the sack. Therefore, there are 9 possible outcomes for the second draw.
To calculate the favorable outcomes, we need to consider the two scenarios where balls of different colors are drawn:
1. First ball is white, and the second ball is red.
There are 4 white balls to choose from for the first draw, and 6 red balls for the second draw. So, there are 4 * 6 = 24 favorable outcomes for this scenario.
2. First ball is red, and the second ball is white.
There are 6 red balls to choose from for the first draw, and 4 white balls for the second draw. So, there are 6 * 4 = 24 favorable outcomes for this scenario.
Adding up the favorable outcomes from both scenarios, we have a total of 24 + 24 = 48 favorable outcomes.
Therefore, the probability of drawing two balls of different colors is 48 (favorable outcomes) divided by 10 (possible outcomes) multiplied by 9 (possible outcomes for the second draw):
Probability = (48 / 10) * (9 / 9) = 48 / 10 = 4.8 / 1 = 4.8
So, the probability is 4.8 or 48%.