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Example: Suppose your football team has 10 returning athletes and 4 new members. How many ways can the coach choose one old player and one new one?
Solution: There are 10 ways to choose a returning player and 4 ways to choose a new one. Thus, there are 10 x 4 = 40 ways to choose one of each.
Example: Suppose we pick 2 numbers from 1, 2, 3, 4, 5 with replacement. Find the probability that one number is even.
Solution: Note that we want only one number to be even, not both, so we can rephrase this as: The first number is even AND the second number is odd OR the first number is odd AND the second number is even.
Use the addition and multiplication rules to calculate that there are (2 x 3) + (3 x 2) = 12 ways for this event to occur. Since there are 5 ways of choosing the first number AND 5 ways of choosing the second, S contains 5 x 5= 25 points.
The probability that one number is even is:
P(one number is even) = 12/25
How does your answer to this question compare with mine, which follows? i) To begin with, 1 laid the beads out in a row for counting, so that I wouldn't leave any out or count a
compare: 643,251: 633,512: 633,893. The answer is 633,512.
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ABCD is a parallelogram in the given figure, AB is divided at P and CD and Q so that AP:PB=3:2 and CQ:QD=4:1. If PQ meets AC at R, prove that AR= 3/7 AC. Ans: ΔAPR ∼ Δ
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