It is known that a box of $550 b u l b s$ contains $22$ defective bulbs. One blubs is taken out at random from the box. Find the probability of getting
(1) defective bulb
(2) good blub

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Solution

$550$ bulbs $(1)$ defective bulb $= \frac{^{22} C_{1}}{^{550} C_{1}} = \frac{22}{550}$
$= \frac{2}{50} 25 = \frac{1}{25}$
$(b) \frac{^{528} C_{1}}{550} = \frac{528}{550}$

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It is known that a box of 550 bulbs contains 22 defective bulbs. One blubs is taken out at random from the box. Find the probability of getting (1) defective bulb(2) good blub

550 bulbs (1) defective bulb =22C1550C1=22550=25025=125(b) 528C1550=528550

It is known that a box of $550 b u l b s$ contains $22$ defective bulbs. One blubs is taken out at random from the box. Find the probability of getting
(1) defective bulb
(2) good blub

$550$ bulbs $(1)$ defective bulb $= \frac{^{22} C_{1}}{^{550} C_{1}} = \frac{22}{550}$
$= \frac{2}{50} 25 = \frac{1}{25}$
$(b) \frac{^{528} C_{1}}{550} = \frac{528}{550}$