A box in a certain supply room contains four 40w, five 60w, and six 75w light-bulbs. suppose that three bulbs are randomly selec
ted. a.what is the probability that exactly two of the selected bulbs are rated 75w?
b.what is the probability that all three of the selected bulbs have the same rating?
c.what is the probability that one bulb of each type is selected?
d.suppose, now, that bulbs are to be selected one by one until a 75w bulb is found. what is the probability that it is necessary to examine at least six bulbs?
Situation satisfies the criteria for the use of hypergeometric distribution. Since no replacement is made, binomial distribution is not applicable (probability does not remain constant).
A=number of target wattage bulbs B=number of non-targeted wattage bulbs a=number of target wattage bulbs selected b=number of non-targeted wattage bulbs selected
P(a,b)=C(A,a)*C(B,b)/C(A+B,a+b) where C(n,x)=combination of x items chosen from n=n!/(x!(n-x)!)
For all following problems, A+B=4+5+6=15 a+b=3 (selected)
(b) target wattage = each of the three Probability = sum of probabilities of choosing 3 40,60,75-watt bulbs P(3x40W)+P(3x60W)+P(3x75W) Case (A,B,a,b) 3x40W (4,11,3,0) 3x60W (5,10,3,0) 3x75W(6,9,3,0)
Answer: 68% of light bulbs last between 1765 hours and 1835 hours.
Step-by-step explanation:
The empirical rule states that for a normal distribution, nearly all of the data will fall within three standard deviations of the mean . The empirical rule is further illustrated below
68% of data falls within the first standard deviation from the mean.
95% fall within two standard deviations.
99.7% fall within three standard deviations.
From the information given, the mean is 1800 hours and the standard deviation is 35 hours.
1 standard deviation = 1 × 35 = 35
1800 - 35 = 1765 hours
1800 + 35 = 1835 hours
Therefore, 68% of light bulbs last between 1765 hours and 1835 hours.
