Question

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

Answer 1

Number of Vehicles in parking lot = 10 [ 3 SUV + 7 Trucks]

Probability of an event =
`\frac{\text{Total favorable outcome}}{\text{Total possible outcome}}`

Probability of choosing 7 Vehicles in a parking spot out of [10 vehicles=3 SUVs and 7 trucks] such that it has 2 SUVS and 5 trucks or 3 SUVS and 4 trucks =
`\frac{_(2)^(3)\textrm{C}*_(5)^(7)\textrm{C}+ _(3)^(3)\textrm{C}*_(4)^(7)\textrm{C}}{_(7)^(10)\textrm{C}}=(3*21+35)/(120)=(63+35)/(120)=(98)/(120)=(49)/(60)`

Use this formula to find , C(n,r)=
`(n!)/((n-r)!r!)`

C(3,2)=3, C(7,5)=21, C(7,4)=35,C(3,3)=1,C(10,7)=120