In: Statistics and Probability
a) What is the minimum number of people that must be in a room to guarantee that at least 20 were born on the same day of the week? Assume all days of the week are equally likely.
b) What is the coefficient of x 9 y 11in ( 3 x − 4 y ) 20?
c) What is the probability that in a permutation of the letters {A, B, C, D, E, F, G, H, I, J, K, L}, the strings BAIL and ILK appear?
d) How many decimal strings of length 15 begin with 321 or end with 9768?
a) There are a total of 7 days in a week. As we are given here
that each day is equally likely. Then to compute the minimum number
of people that must be in a room to guarantee that at least 20 were
born on the same day of the week is computed by first taking the
worst case scenario, which is that for each of the 7 days of the
week, we have 19 people being born on each of those 7 days.
Therefore total number of people covered by that:
= 19*7
= 133
Therefore, we would need a minimum of 134 people in the room to guarantee that at least 20 of them were born on the same day of a week.
b) The polynomial here is given as:
(3x - 4y)20
For x9y11, we compute the coefficient here using the binomial expansion formula as:
This is the required coefficient here.
c) Probability that the string made by the letters {A, B, C, D, E, F, G, H, I, J, K, L}, the strings BAIL and ILK appear is computed here as:
As there is only 1 L here, therefore for BAIL and ILK to
appear,
BAILK will have to appear together.
There are a total of 12 letters here but keeping BAILK aside, we would be left with 7 letters and 1 group BAILK
Therefore the probability here is computed as:
= Number of permutation of (7 + 1) groups / Total permutation of 12
letters
= 8! / 12 !
= 1/11880
Therefore 1/11880 is the required probability here.
d) There are a total of 1015 decimal strings possible for length 15 here.
Now total strings that start with 321 is computed here as:
= Total ways to fill the remaining 12 digits
= 1012
Similarly, total strings that end with 9768 is computed here
as:
= 1011
Total strings that start with 321 and end with 9768
= 10(15 - 3 - 4) = 108
Using law of probability additivity, we get here:
n(start with 321 or end with 9768) = n(start with 321) + n(end with
9768) - n(start with 321 and end with 9768)
= 1012 + 1011 - 108
Therefore the probability now is computed here as:
= (1012 + 1011 - 108 ) / 1015 = 0.001 + 0.0001 - 0.0000001
= 0.0010999
Therefore 0.0010999 is the required probability here.