Question 970742: A calculator has 10 number keys and 5 operation keys.
a)if one key is pressed at random, what is the probability that the key pressed is (i)a number key? (ii)an operation key?
b)When 5 keys are pressed at random, what is the probability that (i)exactly 3 operation keys?(ii)at least 3 are operation keys? (iii)at most 2 number keys?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! (10/15) or 2/3 number key
5/15 or 1/3 operation key
5 keys at random
probability of exactly 3 operation keys
(5/15)*(4/14)*(3/13)*(10/12)*(9/11) This is one form. If the number keys were pressed first and then the operation keys
(10/15)*(9/14)*(5/13)*(4/12)*(3/11). Notice that despite the different order, these multiply out to be the same.
It is 5 C 3 or 10 different ways. That 10 is multiplied by the product of one of those.
(5400)*10/360360
=0.15
For at least 3, one may look at 0,1, 2 and subtract from 1. It is easier.
Probability 0 are operation keys
(10/15)*(9/14)*(8/13)*(7/12)*(6/11) only 1 way for this to occur.
Probability that 1 is operation key
(10/15)*(9/14)*(8/13)*(7/12)*(5/11) *5 ways for that to occur.
Probability that 2 are operation keys
(10/15)*(9/14)*(8/13)*(5/12)*(4/11) * 10 different ways to choose 2 items out of 5.
numerator 1=32340 numerator 2=25200*5=126,000 numerator 3=14,400*10=144,000
302340/360360 = 0.839.
1-0.839=0.161 We got what we DIDN'T WANT and subtracted it from 1, because this is dichotomous
At most 2 number keys
probability that 0 number keys
(5/15)*(4/14)*(3/13)*(2/12)*(1/11)
probability that 1 number key
(5/15)*(4/14)*(3/13)*(2/12)*(10/11) *5 different ways for that to occur.
Probability that 2 number keys
(5/15)*(4/14)*(3/13)*(10/12)*(9/11) * 10 different ways for that to occur.
Numerators sum: 120+6000+54000=60120
Probability= 0.167
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