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Sagot :
Answer:
a) 0.0002 = 0.02% probability that all are silver.
b) 0.0033 = 0.33% probability that 5 are red and 2 are black
c) 0.0345 = 3.45% probability that 4 are red and the rest are silver
Step-by-step explanation:
A probability is the number of desired outcomes divided by the number of total outcomes.
In this question, the order in which the cars were selected is not important, so we use the combinations formula.
Combinations formula:
[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
(a) all are silver
Desired outcomes:
7 silver cars, from a set of 9. So
[tex]D = C_{9,7} = \frac{9!}{7!(9-7)!} = 36[/tex]
Total outcomes:
7 cars from a set of 8 + 9 + 5 = 22. So
[tex]T = C_{22,7} = \frac{22!}{7!(22-7)!} = 170544[/tex]
Probability:
[tex]p = \frac{D}{T} = \frac{36}{170544} = 0.0002[/tex]
0.0002 = 0.02% probability that all are silver.
(b) 5 are red and 2 are black
Desired outcomes:
5 red, from a set of 8
2 black, from a set of 5
[tex]D = C_{8,5}*C_{5,2} = \frac{8!}{5!(8-5)!}*\frac{5!}{2!(5-2)!} = 56*10 = 560[/tex]
Total outcomes:
7 cars from a set of 8 + 9 + 5 = 22. So
[tex]T = C_{22,7} = \frac{22!}{7!(22-7)!} = 170544[/tex]
Probability:
[tex]p = \frac{D}{T} = \frac{560}{170544} = 0.0033[/tex]
0.0033 = 0.33% probability that 5 are red and 2 are black
c) 4 are red and the rest are silver
Desired outcomes:
4 red, from a set of 8
3 silver, from a set of 9
[tex]D = C_{8,4}*C_{9,3} = \frac{8!}{4!(8-4)!}*\frac{9!}{3!(9-3)!} = 70*84 = 5880[/tex]
Total outcomes:
7 cars from a set of 8 + 9 + 5 = 22. So
[tex]T = C_{22,7} = \frac{22!}{7!(22-7)!} = 170544[/tex]
Probability:
[tex]p = \frac{D}{T} = \frac{5880}{170544} = 0.0345[/tex]
0.0345 = 3.45% probability that 4 are red and the rest are silver
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