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Luden [163]3 years ago

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Thirty percent (30%) of the bulbs in a large box are defective. If 12 bulbs are selected randomly from the box, calculate the pr

AleksandrR [38]

The probability that exactly 6 are defective is 0.0792.

Given:

30% of the bulbs in a large box are defective.

If 12 bulbs are selected randomly from the box.

To find:

The probability that exactly 6 are defective.

Solution:

Probability of defective bulbs is:

$p=30\%$

$p=\dfrac{30}{100}$

$p=0.3$

Probability of non-defective bulbs is:

$q=1-p$

$q=1-0.3$

$q=0.7$

The probability that exactly 6 are defective is:

$P(x=r)=^nC_rp^rq^{n-r}$

$P(x=6)=^{12}C_6(0.3)^6(0.7)^{12-6}$

$P(x=6)=924(0.3)^6(0.7)^{6}$

$P(x=6)=0.0792478958$

$P(x=6)\approx 0.0792$

Therefore, the probability that exactly 6 are defective is 0.0792.

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