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Answered

A box contains four 40-W bulbs, five 60-W

bulbs, and six 75-W bulbs. If bulbs are se-

lected one by one in random order, what is

the probability that at least two bulbs must be

selected to obtain one that is rated 75W?

Verified Answer :

syed514

so first one is not a 75W bulb:

P(not a 75Wbulb)=(4+5)/(4+5+6)=9/15

P(a 75W bulb)=6/(15-1)=6/14

9/15*6/14=9/35

JeanaShupp

Answer: 0.99483

Step-by-step explanation:

Given : A box contains four 40-W bulbs, five 60-W  bulbs, and six 75-W bulbs.

Total bulbs : 4+5+6=15

The probability of selecting a 75-W bulb :\(p=\dfrac{6}{15}=0.4\)

Using the binomial probability :-

\(P(X)= ^nC_xp^x(1-p)^{n-x}\), where P(x) is the probability of getting success in x trials , p is the probability of getting success in each trial and n is the total number of trials.

We have,

The  probability that at least two bulbs must be  selected to obtain one that is rated 75W :-

\(P(x\geq2)=1-(P(x<2))\\\\=1-(P(x=0)+P(x=1))\\\\=1-(^{15}C_0(0.4)^0(0.6)^{15}+^{15}C_{1}(0.4)^1(0.6)^{14})\\\\=1-((1)(0.6)^{15}+(15)(0.4)(0.6)^{14})\ \ [\because\ ^nC_0=1\ \&\ ^nC_1=n]\\\\\approx1-(0.00047+0.00470)\\\\=1-0.00517=0.99483\)

Hence, the required probability = 0.99483

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