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There are 10 vehicles in a parking lot: 3 SUVs and 7 trucks. What is the probability that any 7 randomly chosen parking spots have 2 SUVs and 5 trucks OR 3 SUVs and 4 trucks?

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W0lf93

We have to calculate the probability that any randomly choosen parking spots have 2 SUVs and 5 trucks or 3 SUVs and 4 trucks. We will use the formula with combinations: P = ( 3C2 7C5 ) / ( 10C7) + ( 3C3 4C7) / ( 10C7). For example: 10C7 = 10! / 7!(10-7)! = 10!/7!3!= 1098/321 = 120, etc. P = ( 3 21 ) / 120 + ( 1 35 ) / 120 = 63 / 120 + 35 / 120 = 98 / 120 = 0.8166. Answer: The probability is 0.8166 or 81.66%.

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