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The bag contains eight blue marbles, five red marbles, two green marbles, and one black marble. The probability of randomly picking a blue marble is 8/16. What is the probability of not drawing a blue marble? The probability of not picking a blue marble would be 8/16 or the sum of the remaining marbles in the bag.

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Question 1050658: A bag contains 9 red marbles, 8 white marbles, and 6 blue marbles. Randomly choose two marbles, one at a time, and without replacement. Find the following: a) The probability that the first marble is red and the second is white. b) The probability that both are the same color. c) The probability that the second marble is blue.

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You have three jars containing marbles, as follows: jar $1 \quad 600$ red and 400 white jar $2 \quad 900$ blue and 100 white jar $3 \quad 10$ green and 990 white a. If you blindly select one marble from each jar, calculate the probability of obtaining (1) a red, a blue, and a green. (2) three whites. (3) a red, a green, and a white.

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3. If two marbles are chosen at random without replacement, P(they are both red) B. Soda Scenario: You have a cooler full of drinks. There are 12 regular cokes, 6 diet cokes, and 6 coke zeros. Find each of the following probabilities. 1. Find the probability of picking a diet coke. 2.

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What is the probability that the 2 students have the same birthday? Solution: Let the event wherein 2 students having the same birthday be E. Given, P(E) = 0.992. We know, P(E)+P(not E) = 1. Or, P(not E) = 1–0.992 = 0.008 ∴ The probability that the 2 students have the same birthday is 0.008. 8. A bag contains 3 red balls and 5 black balls.

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Mar 28, 2019 · Number of ways of selecting either two green marbles or two yellow marbles = 2 C 2 + 3 C 2 = 1 + 3 = 4 Number of ways of selecting 2 marbles = 15 C 2 = 105 Required Probability [latex]= \frac { 4 } { 105 }[/latex] 6. (d) Total possible outcomes = Number of ways of picking 3 marbles out of 12 marbles = n(S)

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