Question
A bag contains 6 red, 10 yellow, and 14 black balls. If
2 balls are picked at random without replacement, find the probability that the first ball is red and the second ball is either yellow or black.Solution
We have a bag with: 6 Red balls 10 Yellow balls 14 Black balls Total balls = 30 We pick two balls without replacement, and we need the probability that: First ball is Red. Second ball is either Yellow or Black. Step 1: Probability of picking a Red ball first First ball is Red. Second ball is either Yellow or Black. Probability of picking a Red ball first P(R1) = 6/30 = 1/5 Probability of picking a Yellow or Black ball next After picking a red ball, 29 balls remain. Yellow + Black balls left = 10+14 = 24. P(Y2 or B2) = 24/29 Multiply the probabilities P(R1 and (Y2 or B2)) = 1/5 × 24/29 = 24/145
What approximate value will replace the question mark (?) in the following?
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What approximate value will come in place of the question mark (?) in the following question? (Note: You are not expected to calculate the exact value.)...
(70.03 ÷ 3.03 × 12.02) ÷ 35.03 × 20.02 × 8.08 = ? × (9.09 2.02 – 1.01)Â
1242.12 ÷ √530 + 1139.89 ÷ 14.91 = ? + 45.39
27.27% of 5501.22 + 12.53% of 158.99 – √ 1599 = ?
35.1% of 1599 = ?–(449.96 ÷ 6.12) × 2
- What approximate value will come in place of the question mark (?) in the following question? (Note: You are not expected to calculate the exact value.)
(√360.99 + 161.14) ÷ 5 × 249.98 = ?
The greatest number that will divide 398,436, and 542 leaving 7, 11, and 15 as remainders, respectively, is:
- What approximate value will come in place of the question mark (?) in the following question? (Note: You are not expected to calculate the exact value.)