an urn contains 3 blue chips and 7 red chips. sequentially draw 3 chips without replacement. what is the probability that all drawn chips will not be the same color?

prob ( that they are the same colour)

= BBB or RRR
= (3/10)(2/9)(1/8) + (7/10)(6/9)(5/8)
= 6/720 + 210/720
= 216/720
= 3/10

prob(not the same colour) = 1 - 3/10 = 7/10

Many thanks!!!

To find the probability of drawing three chips without replacement and having them all be different colors, we need to calculate the probability of each possible outcome and add them together.

First, let's find the probability of drawing one blue chip and two red chips. The probability of drawing a blue chip on the first draw is 3/10 (since there are 3 blue chips out of 10 total chips). After drawing a blue chip, there are 9 remaining chips, 7 of which are red. The probability of drawing a red chip on the second draw is 7/9. After drawing a red chip on the second draw, there are 8 remaining chips, 6 of which are red. The probability of drawing a red chip on the third draw is 6/8. So, the probability of drawing one blue chip and two red chips is (3/10) * (7/9) * (6/8) = 0.175.

Next, let's find the probability of drawing one red chip and two blue chips. The probability of drawing a red chip on the first draw is 7/10. After drawing a red chip, there are 9 remaining chips, 3 of which are blue. The probability of drawing a blue chip on the second draw is 3/9. After drawing a blue chip on the second draw, there are 8 remaining chips, 2 of which are blue. The probability of drawing a blue chip on the third draw is 2/8. So, the probability of drawing one red chip and two blue chips is (7/10) * (3/9) * (2/8) = 0.07.

Finally, we add the probabilities of the two cases to get the probability of drawing three chips without replacement and having them all be different colors: 0.175 + 0.07 = 0.245.

Therefore, the probability that all drawn chips will not be the same color is 0.245, or 24.5%.