Question 1172133: Two Laptop salesmen A and B must each make two calls per day, one in the morning and in the afternoon. A has a probability 0.4 of selling a laptop on call, while B has a probability 0.1 of on a call. A works independently of B, and for each salesman, morning and afternoon are independent of each other.
Find the probability that, in one day:
A sells two laptops [1 Marks]
A sells just one laptop [ 1 Marks]
B makes at least one sale [ 2 Marks]
Between them A and B make exactly one sale. [3 Marks]
Found 2 solutions by 69420, Theo:
Answer by 69420(3)   (Show Source):
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! A probability of selling a laptop on a call is .4
B probability of selling a laptop on a call is .1
A probability of not selling a laptop on a call is 1 - .4 = .6
B probability of not selling a laptop on a call is 1 - .1 = .9
when two events are independent, the probability of both events occurring is equal to the probability of the first event occurring times the probability of the second event occurring.
the probability that A sells two laptops in one day is equal to .4 * .4 = .16.
the probability that A sells just 1 laptop in one day is equal to .4 * .6 + .6 * .4 = .24 + .24 = .48
since the events are independent, the probability of each event occurring is the same in both events.
the probability that B makes at least one sale is equal to 1 minus the probability that B makes 0 sales.
the probability that B makes 0 sales is equal to .9 * .9 = .81.
the probability that B makes at least 1 sale is equal to 1 - .81 = .19.
in order for A and B to make exactly one sale, either A makes 1 sale or B makes 1 sale but both do not make 1 sale.
the probability that A makes exactly 1 sale is .4 * .6 + .6 * .4 = .24 + .24 = .48
the probability that A makes 0 sales is equal to .6 * .6 = .36
the probability that B makes exactly 1 sale is .1 * .9 + .9 * .1 = .09 + .09 = .18.
the probability that B makes 0 sales is equal to .9 * .9 = .81
the probability that A makes exactly 1 sale and B makes exactly 0 sales is equal to .48 * .81 = .3888.
the probability that A makes exactly 0 sales and B makes exactly 1 sale is equal to .36 * .18 = .0648
the probability that A makes exactly 1 sale and B makes exactly 0 sales or A makes exactly 0 sales and B makes exactly 1 sale is equal to .3888 + .0648 = .4536.
as in all probability type problems, the sum of all possible probabilities must be equal to 1.
for A, this becomes:
p(0 sales) = .6 * .6 = .36
p(1 sale) = .4 * .6 + .6 * .4 = .24 + .24 = .48
p(2 sales) = .4 * .4 = .16
the sum of all the probabilities is equal to 1, as it should.
for B, this becomes:
p(0 sales) = .9 * .9 = .81
p(1 sale) = .1 * .9 + .9 * .1 = .18
p(2 sales) = .1 * .1 = .01
sum of all probabilities is equal to 1, as it should.
for A and B,....
p(0 sales for both) = .36 * .81 = .2916
p(0 sales for A and 1 sale for B) = .36 * .18 = .0648
p(0 sales for A and 2 sales for B) = .36 * .01 = .0036
p(1 sale for A and 0 sales for B) = .48 * .81 = .3888
p(1 sale for A and 1 sale for B) = .48 * 18 = .0864
p(1 sale for A and 2 sales for B) = .48 * .01 = .0048
p(2 sales for A and 0 sales for B) = .16 * .81 = .1296
p(2 sales for A and 1 sale for B) = .16 * .18 = .0288
p(2 sales for A and 2 sales for B) = .16 * .01 = .0016
sum of all probabilities is equal to 1, as it should.
it looks like the probabilities have been calculated correctly since the sum of all possible probabilities for A, and for B, and for A and B are all equal to 1, as they should be.
keep in mind, that if the events are independent of each other, then the probabilities for each event don't change.
if the events are dependent on each other, then the probabilities will change.
this problem made it clear that all events were independent of each other, so the second event is not dependent on the results of the first event, as it would be if the events were not independent of each other.
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