College Math

College Math Study Guide

6. c Solution: We make use of distinguishable permutation in this case. The total number of letters are 9, thus, n = 9. But letter P is repeated thrice, I is written once, N is written once, E is repeated twice, A is written once and L is written once. The total number of ways becomes: ! 1 ! ∗ 2 ! ∗ 3 ! ∗ 4 ! ∗ 5 ! ∗ 6 ! 9! 3! ∗ 1! ∗ 1! ∗ 2! ∗ 1! ∗ 1! = 30,240 7. a Solution: The total sum on two dice can range between 2 and 12. The total number of elements in the sample space would be calculated by the counting principle. The total possibilities on each die is 6, hence n = 6*6 = 36. Now the event = sum on two dice is 8 is possible in the following cases: Dice 1 2 3 4 5 6 Dice 2 6 5 4 3 2 Hence, n(E) = 5 and n(S) = 36 P(E) = 5/36 = 0.1389 =13.89% 8. b Solution: This is the case of conditional probability. Let the draw of a red marble be P(A) and the drawing of a blue marble be P(B). It is given that P(A) = 0.47 and P(A∩B) = 0.32. Now we have to find P(B|A). ( | ) = ( ∩ ) ( ) = 0.32 0.47 = 0.6808 = 68%

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