Respuesta :
If the order does not matter, cars can be arranged in 20 ways
If an order is important, cars can be arranged in 120 ways
The probability that the three newest cars end up parked in the driveway is 0.167
1. If the order does not matter, combination is used
[tex]\left ({n} \atop {r}} \right.)=\frac{n!}{r!(n-r)!}[/tex]
Here, n=6 r=3
using formula,we get
[tex]\left ( {{6} \atop {3}} \right)=\frac{6!}{3!3!} =5*4=20[/tex]
2. If an order is important, Permutation will be applicable
[tex]^{n}P_{r} = \frac{n!}{(n-r)!} \\[/tex]
∴[tex]^{6} P_{3} =\frac{6!}{3!(6-3)!}=\frac{6!}{3!} =120[/tex]
3. the probability that the three newest cars end up parked in the driveway
P=[tex]\frac{No. of possible outcomes for 3 cars}{total possibilities}[/tex]
=[tex]\frac{6*4*5}{6!}[/tex]
=[tex]\frac{120}{720}[/tex]
=[tex]\frac{1}{6}[/tex] ≈ 0.167
Hence, If the order does not matter, cars can be arranged in 20 ways
If an order is important, cars can be arranged in 120 ways
The probability that the three newest cars end up parked in the driveway is 0.167
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