The probability of getting two good coils when two coils are randomly selected if the first selection is replaced before the second is made is 0.7744

<u>Solution:</u>

Total number of coils = number of good coils + defective coils = 88 + 12 = 100

p(getting two good coils for two selection) = p( getting 2 good coils for first selection ) $\times$ p(getting 2 good coils for second selection)

p(first selection) = p(second selection) = $\frac{\text { number of good coils }}{\text { total number of coils }}$

Hence, p(getting 2 good coil for two selection) = $\frac{88}{100} \times \frac{88}{100} =\bold{0.7744}$

5 0

3 years ago

Please Help Part C Final Part Thanks

alexandr402 [8]

3 blue + 3 yellow = 6 total marbles.

The probability of picking a blue one first would be 3/6 ( 3 blue out of 6 total marbles).

3/6 reduces to 1/2.

After picking the first marble, there would be 5 marbles left in the bag.

The probability of picking a yellow one would be 3/5 ( 3 yellow and 5 total marbles left).

The probability of picking a blue then a yellow then becomes:

1/2 x 3/5 = 3/10 = 0.30 = 30%

4 0

3 years ago

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