183k views
5 votes
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?

2 Answers

5 votes

Answer: 0.99483

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=(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\ \&amp;\ ^nC_1=n]\\\\\approx1-(0.00047+0.00470)\\\\=1-0.00517=0.99483

Hence, the required probability = 0.99483

User Jonypz
by
6.7k points
2 votes
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
User Juffy
by
7.2k points