Question Bank  Modern Math  Hemant Malhotra

Q100) The probability that a man who is 85 year old will die before attending the age of 90 year is 1/3.
A1, A2, A3 and A4 are four persons who are 85 year old. The probability that A1 will die before attending the age of 90 year and will be the first to die is?

@hemant_malhotra (4^23^2)(3^22^2)(5^24^2)

@hemant_malhotra
a + 5 + b + 5 + c + 1 + d + 1 = 25
a + b + c + d = 13
16c3 = 560

@hemant_malhotra 960 ways

@hemant_malhotra 115

@hemant_malhotra 4c2 * 4c3 = 24 ways

@hemant_malhotra 17/100

@hemant_malhotra 10C4 * 6c3 * 3c2 * 1c1

Total 9 letters
a  3 times
b  2 times
c  4 times
so total arrangements = 9!/(3! * 2! * 4!) = 1260

@hemant_malhotra 15 numbers

@badalravi Correct!

@ritesh9 Correct!

@badalravi
Total 30 students in 3 sections. So the selection can be done in 30C3 ways.
As we cannot have all 3 from the same section, we should remove those cases.
All 3 from Section A = 8C3
All 3 from Section B = 10C3
All 3 from Section C = 12C3
So possible ways = 30C3  (8C3 + 10C3 + 12C3) = 4060  396 = 3664 ways

@hemant_malhotra its 40% > 10
So 80% > 20

@badalravi
n(A) = 60
n(B) = 120
n(C) = 105
None = 60
Only A = 30
B & A = 30It's easy to draw venn diagram. Let me just type! :D
As n(A) = 60, n(B&A) = 30 and Only A = 30 > We can say n(A&B&C) = 0 and n(A&C) = 0n(AUBUC) = n(A) + n(B) + n(C)  (n(A&B) + n(A&C) + n(B&C)) + n(A&B&C)
240 = 60 + 120 + 105  (30 + 0 + n(B&C)) + 0
240 = 285  30  n(B&C)
n(B&C) = 15
So Only C = 105  15 = 90

@hemant_malhotra total students = 580
girls = 348

@hemant_malhotra 6!*2!*9C4

@hemant_malhotra
222=8

@hemant_malhotra
I+2II +3III = 240
I+II+III = 50
II =18020=160

@hemant_malhotra
max = 5