A box contains 100bulbs, 20 of which are defective. 10 bulbs are selected for inspection. Find the probability that:

(i) all 10 are defective

(ii) all 10 are good

(iii) at least one is defective

(iv) none is defective

Asked by Aaryan | 1 year ago |  34

##### Solution :-

Given: A box contains 100bulbs, 20 of which are defective.

By using the formula,

P (E) = $$\dfrac{favourable\; outcomes }{total\; possible\; outcomes}$$

Ten bulbs are drawn at random for inspection,

Total possible outcomes are 100C10

n (S) = 100C10

(i) Let E be the event that all ten bulbs are defective

n (E) = 20C10

P (E) = $$\dfrac{ n (E) }{ n (S)}$$

$$\dfrac{ ^{20}{C}_{10}}{^{100}C_{10}}$$

(ii) Let E be the event that all ten good bulbs are selected

n (E) = 80C10

P (E) =$$\dfrac{ n (E) }{ n (S)}$$

$$\dfrac{^{ 80}C_{10}} {^{ 100}C_{10}}$$

(iii) Let E be the event that at least one bulb is defective

E= {1,2,3,4,5,6,7,8,9,10} where 1,2,3,4,5,6,7,8,9,10 are the number of defective bulbs

Let E′ be the event that none of the bulb is defective

n (E′) = 80C10

P (E′) = $$\dfrac{ n (E') }{ n (S)}$$

$$\dfrac{^{ 80}C_{10}} {^{ 100}C_{10}}$$

So, P (E) = 1 – P (E′)

=$$1-\dfrac{^{ 80}C_{10}} {^{ 100}C_{10}}$$

(iv) Let E be the event that none of the selected bulb is defective

n (E) = 80C10

P (E) = $$\dfrac{ n (E) }{ n (S)}$$

$$\dfrac{^{ 80}C_{10}} {^{ 100}C_{10}}$$

Answered by Aaryan | 1 year ago

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