Question

The numbers 1 to 15 are written on 15 pieces of paper and dropped into a box. Three of
them are drawn at random. What is the probability that the three pieces of paper picked
have numbers that are in arithmetic progression?
0 1.508
O 0.107
1.058
O 0.564​

Answer 1

Answer:

total sample space={1,2,3,4,5,6,7,8,9,10,11}

Total number of cases=3 pieces selected from 11 pieces in 11c3 ways=11!/(3!*8!)=165

probability= number o favourable cases/total number of cases.

favourable cases;

if d=1

favourable cases=(1,2,3),(2,3,4),(3,4,5),(4,5,6),(5,6,7),(6,7,8),(7,8,9),(8,9,10),(9,10,11)

d=2

favourable cases=1,3,5),(2,4,6),(3,5,7),(4,6,8),(5,7,9),(6,8,10),(7,9,11)

if d=3

favourable cases =(1,4,7),(2,5,8),(3,6,9)(4,7,10),(5,8,11)

if d=4

favourable cases =(1,5,9),(2,6,10),(3,7,11)

if d=5

favourable cases=(1,6,11)

so total number of favourable cases=9+7+5+3+1=25

probability=25/165

Answer 2

Step-by-step explanation:

Soooo the answer is 0.158