Asample of 3 different bulbs is randomly selected from a group containing 8 that are defective and 22 that have no defects. what is the probability that at least one of the bulbs is defective? round answer to three decimal places.

Probability of drawing non defective bulb

=90/100×89/99×88/98××82/92(i)=90/100×89/99×88/98××82/92(i)

Now the probability of drawing at least one bulb will be defective

=1−(i)=1−90/100×89/99×88/98××82/92

1−(90/100⋅89/99⋅…⋅83/93)

how am i supposed to answer

step-by-step explanation:

1. 3. can also be expressed as (22/7)

2. -4.5

3. 3.6

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