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)
In this problem, you apply principles in trigonometry. Since it is not mentioned, you will not assume that the triangle is a special triangle such as the right triangle. Hence, you cannot use Pythagorean formulas. The only equations you can use is the Law of Sines and Law of Cosines.
For finding side a, you can answer this easily by the Law of Cosines. The equation is
a2=b2 +c2 -2bccosA a2 = 11^2 + 8^2 -2(11)(8)(cos54) a2 = 81.55 a = √81.55 a = 9
Then, we use the Law of Sines to find angles B and C. The formula would be
The fraction (1 yard / 36 inches) has a numerator and denominator that are equal. So the fraction is equal to ' 1 ', and we can multiply (26 inches) by it without changing any value. That'll change the units of the '26 inches' to 'yards', which is exactly what we need.
