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!= 10*9*8/3*2*1 = 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%.
