AP EAPCET 18 May 2024 Shift 1 Solved Paper

To solve for the probability that exactly 3 out of 5 bulbs chosen are defective, given that 20% of the bulbs are defective, we can use the binomial probability formula. Here's how it's done:
Define the variables:
Total number of trials, n=5
Probability of a defective bulb, p=20%=‌

1

5

Probability of a non-defective bulb, q=1−p=1−‌

1

5

=‌

4

5

Calculate the probability:
We need to find the probability of exactly 3 defective bulbs, P(X=3).
Apply the binomial probability formula:
P(X=3)=‌5C3(‌

1

5

)3(‌

4

5

)2
Calculate the individual components:
The coefficient, ‌5C3, is the number of ways to choose 3 defective bulbs from 5 . This is calculated as:
‌5C3=‌

5!

3!(5−3)!

=‌

5×4×3!

3!×2×1

=10
Plug in the values:
‌(‌

1

5

)3=‌

1

125

‌(‌

4

5

)2=‌

16

25

Calculate the probability:
P(X=3)=10×‌

1

125

×‌

16

25

=10×‌

16

3125

=‌

160

3125

=‌

32

625

Therefore, the probability that exactly 3 out of the 5 bulbs chosen are defective is ‌

32

625

.