Probability 1
A bag contains tickets numbered from 1 to 2 5 . Five tickets are drawn at random and arranged in ascending order. Find the probability that 2 n d and 5 t h bear the numbers 1 2 and 1 8 respectively.
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3 solutions
here is the answer
the total probability of getting 5 cards out of the box is (25 C 5) which infact becomes our denominator.
The numerator is
selecting 12 on 2nd is ( 25 C 1) * (1/5) as 12 can be any of the 5 tickets taken out thats why 1/5. .............. - (1)
Similarly selecting 18 on 2nd is ( 25 C 1) * (1/5) as 12 can be any of the 5 tickets taken out thats why 1/5.......................................... - (2)
so u notice it is & condition so we multiply (1) and (2 ) to get numerator.
answer is 1/53130
Well the answer should be 1/26565
How many of you agree ??
@manish bhargao Can you please post solution for it
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I s t h i s c o r r e c t ? 1 s t p o s i t i o n − 1 1 c h o i c e s 2 n d p o s i t i o n − 1 c h o i c e ( 1 2 ) 3 r d p o s i t i o n − 1 3 , 4 t h p o s i t i o n − 4 c h o i c e s 3 r d p o s i t i o n − 1 4 , 4 t h p o s i t i o n − 3 c h o i c e s 3 r d p o s i t i o n − 1 5 , 4 t h p o s i t i o n − 2 c h o i c e s 3 r d p o s i t i o n − 1 6 , 4 t h p o s i t i o n − 1 c h o i c e T o t a l = 1 0 c h o i c e s 5 t h p o s i t i o n − 1 c h o i c e ( 1 8 ) . T o t a l n u m b e r o f s o l u t i o n s = 1 1 ∗ 1 0 = 1 1 0 T o t a l n u m b e r o f c h o i c e s = C ( 2 5 , 5 ) . P ( E ) = 1 / 4 8 3 .
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I completely agree ....did exact same
Why are there only 11 choices in 1st positions...
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@Manish Bhargao – since arrangement is in ascending order and 2nd ticket is 12.
Completely agree!
Adarsh Kumar the answer is 1/138
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@Utkarsh Bansal – What is wrong with this?
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@Adarsh Kumar – Even I agree. t he answer is 1/483
Well I am not at all sure about my solution.... i just gave this an attempt.... n(S)=25C5*1......(since there is only one way to arrange them i.e in ascending order) then I chose five places and started putting the tickets randomly...which can be done in the way... @Adarsh Kumar stated
P.S My answer is wrong...... Also please post your solution...
Here's a short C code I used to calculate the probability.
If
m
=
6
0
is the number of strings where
1
2
and
1
8
are fixed and
m
=
4
2
5
0
4
the number of all ascending strings, than
P = 4 2 5 0 4 6 0 = 3 5 4 2 5