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Sagot :
Answer:
a.) 0.5
b.) 0.66
c.) 0.83
Step-by-step explanation:
As given,
Total Number of Batteries in the drawer = 10
Total Number of defective Batteries in the drawer = 4
⇒Total Number of non - defective Batteries in the drawer = 10 - 4 = 6
Now,
As, a sample of 3 is taken at random without replacement.
a.)
Getting exactly one defective battery means -
1 - from defective battery
2 - from non-defective battery
So,
Getting exactly 1 defective battery = ⁴C₁ × ⁶C₂ = [tex]\frac{4!}{1! (4 - 1 )!}[/tex] × [tex]\frac{6!}{2! (6 - 2 )!}[/tex]
= [tex]\frac{4!}{(3)!}[/tex] × [tex]\frac{6!}{2! (4)!}[/tex]
= [tex]\frac{4.3!}{(3)!}[/tex] × [tex]\frac{6.5.4!}{2! (4)!}[/tex]
= [tex]4[/tex] × [tex]\frac{6.5}{2.1! }[/tex]
= [tex]4[/tex] × [tex]15[/tex] = 60
Total Number of possibility = ¹⁰C₃ = [tex]\frac{10!}{3! (10-3)!}[/tex]
= [tex]\frac{10!}{3! (7)!}[/tex]
= [tex]\frac{10.9.8.7!}{3! (7)!}[/tex]
= [tex]\frac{10.9.8}{3.2.1!}[/tex]
= 120
So, probability = [tex]\frac{60}{120} = \frac{1}{2} = 0.5[/tex]
b.)
at most one defective battery :
⇒either the defective battery is 1 or 0
If the defective battery is 1 , then 2 non defective
Possibility = ⁴C₁ × ⁶C₂ = 60
If the defective battery is 0 , then 3 non defective
Possibility = ⁴C₀ × ⁶C₃
= [tex]\frac{4!}{0! (4 - 0)!}[/tex] × [tex]\frac{6!}{3! (6 - 3)!}[/tex]
= [tex]\frac{4!}{(4)!}[/tex] × [tex]\frac{6!}{3! (3)!}[/tex]
= 1 × [tex]\frac{6.5.4.3!}{3.2.1! (3)!}[/tex]
= 1× [tex]\frac{6.5.4}{3.2.1! }[/tex]
= 1 × 20 = 20
getting at most 1 defective battery = 60 + 20 = 80
Probability = [tex]\frac{80}{120} = \frac{8}{12} = 0.66[/tex]
c.)
at least one defective battery :
⇒either the defective battery is 1 or 2 or 3
If the defective battery is 1 , then 2 non defective
Possibility = ⁴C₁ × ⁶C₂ = 60
If the defective battery is 2 , then 1 non defective
Possibility = ⁴C₂ × ⁶C₁
= [tex]\frac{4!}{2! (4 - 2)!}[/tex] × [tex]\frac{6!}{1! (6 - 1)!}[/tex]
= [tex]\frac{4!}{2! (2)!}[/tex] × [tex]\frac{6!}{1! (5)!}[/tex]
= [tex]\frac{4.3.2!}{2! (2)!}[/tex] × [tex]\frac{6.5!}{1! (5)!}[/tex]
= [tex]\frac{4.3}{2.1!}[/tex] × [tex]\frac{6}{1}[/tex]
= 6 × 6 = 36
If the defective battery is 3 , then 0 non defective
Possibility = ⁴C₃ × ⁶C₀
= [tex]\frac{4!}{3! (4 - 3)!}[/tex] × [tex]\frac{6!}{0! (6 - 0)!}[/tex]
= [tex]\frac{4!}{3! (1)!}[/tex] × [tex]\frac{6!}{(6)!}[/tex]
= [tex]\frac{4.3!}{3!}[/tex] × 1
= 4×1 = 4
getting at most 1 defective battery = 60 + 36 + 4 = 100
Probability = [tex]\frac{100}{120} = \frac{10}{12} = 0.83[/tex]
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