A book has 411 pages. The reader flips it open to a random page. What is the probability that the page number it is open to is divisible neither by 2 nor 3?
Answers
Answer:
7 could be the right answer
Given : book has 411 pages. The reader flips it open to a random page.
To find : What is the probability that the page number it is open to is divisible neither by 2 nor 3?
Solution:
Page number open is neither Divisible by 2 nor by 3
= total pages - { (pages divisible by 2) ∪ (Pages divisible by 3) }
(pages divisible by 2) ∪ (Pages divisible by 3)
= (pages divisible by 2) + (Pages divisible by 3) - { (pages divisible by 2) ∩ (Pages divisible by 3) }
(pages divisible by 2) ∩ (Pages divisible by 3) = Pages divisible by LCM of 2 & 3 = 6
pages divisible by 2 = 2 , 4 .............................410
= 2( 1 , 2 , ...............................205)
205 pages
pages divisible by 3 = 3 , 6 .............................411
= 3( 1 , 2 ........................137 )
137 pages
pages divisible by 12 = 6 , 12 .............................408
= 6 ( 1 , 2 .................................. 68 )
68 Pages
(pages divisible by 2) ∪ (Pages divisible by 3) = 205 + 137 - 68
= 274 Pages
Page number open is neither Divisible by 2 nor by 3 = 411 - 274
= 137
Probability Page number open is neither Divisible by 2 nor by 3 = 137/411
= 1/3
1/3 is the probability that page number it is open is divisible neither by 2 nor 3
Another simpler method
Break 1 to 411 into 137 Group
( 1 , 2 , 3) , ( 4 , 5 , 6 ) ........................................... (409 ,410 , 411)
Each group has only one number which is neither divisible by 2 , nor by 3
Hence probability = 1/3
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