In a box contain 4 green balls 4 white balls and 4 red balls .If 4 balls are drown randomly what is the probability that no one even out of the 4 balls being white.
"that no one even out of the 4 balls being white" ??
you mean, none of the balls drawn is white ?
8 of the 12 are "non-white"
prob(of your event)
= (8/12)(7/11)(6/10)(5/9)
= 14/99
assuming the balls are not returned after they are drawn
To find the probability of not drawing any white balls out of four draws, we need to calculate two probabilities separately:
1. The probability of drawing a non-white ball on the first draw.
2. The probability of drawing a non-white ball on the second draw, given that the first draw was a non-white ball.
Assuming that each ball has an equal chance of being drawn, let's calculate these probabilities step by step:
1. The probability of drawing a non-white ball on the first draw:
- There are a total of 12 balls in the box (4 green + 4 white + 4 red).
- Since we don't want to draw a white ball, there are 8 non-white balls left in the box (4 green + 4 red).
- Therefore, the probability of drawing a non-white ball on the first draw is 8/12 or 2/3.
2. The probability of drawing a non-white ball on the second draw, given that the first draw was a non-white ball:
- After the first draw, there are 11 balls left in the box.
- If the first draw was a non-white ball, there are still 8 non-white balls left (4 green + 4 red) out of the 11 balls.
- Therefore, the probability of drawing a non-white ball on the second draw, given that the first draw was a non-white ball, is 8/11.
To calculate the overall probability of not drawing any white balls on four draws, we need to multiply the probabilities of each individual draw since these events are independent:
(2/3) * (8/11) * (8/10) * (7/9) ≈ 0.2269
So, the probability of not even drawing a single white ball out of four draws is approximately 0.2269 or 22.69%.