Probability with selecting colored balls?
Rework problem 28 from section 3.3 of your text, involving the selection of colored balls from a box. Assume that the box contains 10 balls: 3 red, 3 blue, and 4 yellow. As in the text, you draw one ball, note its color, and if it is yellow replace it. If it is not yellow you do not replace it. You then draw a second ball and note its color. (1) What is the probability that the second ball drawn is yellow? (2) What is the probability that the second ball drawn is red?
Probability - three urns, coloured balls - please help?
probability of selecting any one urn from 3 urns is (1/3) Probability of selecting urn 1 and then a red ball from urn 1 is (1/4) [3C1/(3+4+1) C1] = (1/3)(3/8) Probability of selecting urn 2 and then a red ball from urn 2 is (1/4) [1C1/(1+ 2 +3)C1] = (1/3)(1/6) Probability of selecting urn 3 and then a red ball from urn 3 is (1/4) [4C1/(4+ 3 + 1)C1] = (1/3)(4/9) By Bayes theroem given that the event red ball has taken withdrawn Probabiluity that this red ball is from Urn 1 is [(1/3)(3/8)] / [(1/3)(3/8) + (1/3)(1/6) + (1/3)(4/9)] = 81/213
A bag contains 8 white and 6 red balls. What is the probability of drawing two balls of the same color?
We really need more information, for example, how many balls are you drawing? If it's three or more, the probability is 100%.However, I will assume that you intend to draw only two balls and that you don't replace the first ball in the bag for the second draw.So you draw one ball. The probability that it's red is 8/14 = 4/7.You draw again. The probability of getting a second red is 7 (red balls remains) / 13 (balls remaining).Assuming that the results of the first and second draws are independent,the probability of drawing two red balls is therefore (4/7)*(7/13) = 4/13.Now, suppose instead of a red on the first draw you got white. The probability of this is 6/14 = 3/7.Given the first ball is white, the probability of drawing a second white is 5/13.So the probability of drawing two whites is (3/7)*(5/13) = 15/91.The outcomes of drawing two reds or two whites are independent of each other, so you can add the probabilities to get the probability of drawing two balls of the same colour, i.e.4/13 + 15/91 = 43/91If you prefer to express it as a percentage, it's approximately 47.25%.
What is the probability that the two balls have different colors?
Your calculation is right. As Gregory Schoenmakers stated, you even don’t have to bother with the urn selection probabilities, as the probability of a different colored pair is independent of the urns. Both urns has 2 from one color, and 3 from another, so the probability must be 3/5. It would be the same even with any other urn selection probabilities.
What is the logic behind probability of drawing balls from a bag having multiple color balls?
P (Event) = No. of favorable outcomes / No. of total outcomesLet there be a bag with 10 blue balls, 7 red balls, 16 white balls, 8 green balls and 5 purple balls.Thus no. of total balls = 10 + 7 + 16 + 8 + 5 = 46B = A blue ball is drawnThen P(B) = 10/46 = 5/23So in a nutshell, probability of taking out a ball of color X is the quotient of total balls of that color in the bag to the total balls in the bag