Question

a Flashlight has 10 batteries out of which 4 are dead if 3 batteries are selected without replacement and tested then the probability that all three are dead is

Answer 1

Total number of outcomes = $^{10} C_{3}$ 
(3 batteries to be selected from  a Flashlight of10 batteries)

Total number of possible outcomes = $^{4} C_{3}$
(3 batteries to be selected from a flashlight of 4 dead batteries out of 10)

Probability = $\frac{Total number of possible outcomes}{Total number of outcomes}$
                  =$\frac{[tex] ^{4} C_{3}$}{$^{10} C_{3}$} [/tex]
                  =$\frac{ \frac{4}{1} }{ \frac{10*9*8}{3*2} }$
                  = 1/30

Answer 2

Number of flashlight batteries = 10
There are three batteries are selected out of 10
So, number of possible outcomes { sample space} = ¹⁰C₃ = 10!/3!(10-3)!
= 10!/3!7! = 10 × 9 × 8/6 = 120

There are four batteries are dead out of 10 .
And we have to select three dead batteries out of 4 dead batteries
So, number of favourable outcomes = ⁴C₃ = 4!/3!(4-3)! = 4

Now, probability = number of favourable outcomes/number of possible outcomes
= 4/120 = 1/30