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A bag containing 4 green 6 black 7 white balls a ball is drawn at randomly probability of finding 2 red balls?
The probability is zero, because there are no red balls in the bag.
What is the probability of pulling a black ball and then a green ball if there are 3 red balls 3 green balls 3 blue balls and 3 black balls?
3/12*3/11 = 9/132, or 6.818%.
There are 2 red and 5 green balls in a bag If you randomly choose balls one at a time with replacement what is the probability of choosing 2 green balls and then 1 red ball?
Probability of first green: 5/7Probability of second green: 5/7Probability of a red: 2/7Probability of all 3 consecutive successes: (5/7) x (5/7) x (2/7) = 50/343 = 14.58% (rounded)
There are two identical bottles One bottle contains 2 green balls and 1 red ball The other contains 2 red balls A bottle is selected at random and a single ball is drawn What is the probabilit?
nobodys gonna answer that weird question
What is the probability of drawing a white ball out of a bag with 2 green 3 white and 5 black?
The probability of a simple event is the number of ways to succeed over the total possibilities. So, there are three white balls, and ten balls total. So: P(white) = 3/10
A bag containing 4 green 6 black 7 white balls a ball is drawn at randomly probability of finding 2 red balls?
The probability is zero, because there are no red balls in the bag.
These bag contains 6 white ball 3 red balls and one blue ballif one ball is drawn from the bag what is the probability it will be white?
3/5 probability of a white ball being drawn.
One ball is drawn at random from a bag containing 17 red balls and 4 white balls what is the probability that the ball is red?
17 out of 21
If you have 4 green balls 2 red balls and 1 blue what is probability of getting a green?
It is 1/7 if a ball is selected at random.
What is the probability of pulling a black ball and then a green ball if there are 3 red balls 3 green balls 3 blue balls and 3 black balls?
3/12*3/11 = 9/132, or 6.818%.
If there are 6 green balls 3 red balls and 1 pink ball in a container what is the probability of choosing a green ball without looking in the container first?
6/10 or reduced is 3/5
A bag contain 4 green 6 black 7 white balls a ball is drawn at random what is the probability that it is a green or a black ball?
This is a law of addition probability which states that the probability of A or B equals the probability of A plus the probability of B minus the probability of A and B. Written in mathematical terms, the equation is: P(AorB) = P(A) + P(B) - P(AnB) where P(AnB) = 0 (since you can not pull out a green and black ball at the same time). Let P(A) = Probability of drawing the green ball & let P(B) = Probability of drawing the black ball. Total outcomes is 17. So, P(A) = 4/17 & P(B) = 6/17. Therefore P(green or black) = 4/17 + 6/17 = 10/17.
If there are 4 balls in a bag one is blue and the rest are red what is the probability to pull out a green ball?
zero
What is the probability of picking a red ball out of a bag containing 3 red balls 4 blue balls and 6 green balls?
3 out of 13 or 3/13
When an urn contains 12 red balls 153 blue balls 2 black balls 47 white balls and no green balls What is the probability you do not draw any ball at all?
2
What conclusion can be drawn based on the information in the table of balls'?
Based on the information in the table, it can be concluded that the red ball has the highest weight while the green ball has the lowest weight. The blue ball falls in between the red and green balls in terms of weight.
What is probability that you pick 3 white balls if you have 3 white balls 1 red ball and 2 green balls?
The probability that you pick 3 white balls is 0.05 or 5%.----------------------------------------------------------------------------------------------------EXPLANATIONThe probability of first drawing a white ball is: P(W1) =3/6.The probability of drawing a white ball given the event that you already draw awhite ball in the first draw and not replacing it back is: P(W2│W1) =2/5.The probability of drawing a white ball on the third draw given the event that awhite ball was drawn in the first and in the second draw is: P(W3│(W1UW2)) =1/4Now, the probability of drawing 3 white balls one by one with out replacement(taking all 3 balls at a time gets the same analysis and result) is:P(W1UW2UW3) =P(W1)∙P(W2│W1)∙P(W3│(W1UW2)) =(3/6)∙(2/5)∙(1/4) =1/20=0.05 =5%.
