Since PQ bisects both MO and NO, it should be exactly 1/2 of the far parallel side.
PQ = 1/2 MN
PQ = 1/2×14.4 = 7.2
Procedure:
1) calculate the number of diferent teams of four members that can be formed (with the ten persons)
2) calculate the number of teams tha meet the specification (two girls and two boys)
3) Divide the positive events by the total number of events: this is the result of 2) by the result in 1)
Solution
1) the number of teams of four members that can be formed are:
10*9*8*7 / (4*3*2*1) = 210
2) Number of different teams with 2 boys and 2 girls = ways of chosing 2 boys * ways of chosing 2 girls
Ways of chosing 2 boys = 6*5/2 = 15
Ways of chosing 2 girls = 4*3/2 = 6
Number of different teams with 2 boys and 2 girls = 15 * 6 = 90
3) probability of choosing one of the 90 teams formed by 2 boys and 2 girls:
90/210 = 3/7
IDK. I looked up the answer but i couldn't find it. I'm on UsaTestPrep and they gave me the same question.
