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Probability question
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sanjayr84 sanjayr84 is offline
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Re: Probability question - 01-08-2007, 04:07 PM

case 1: king of spade and a king
(1/52)*(3/51)

Case 2: non-king spade and king
(12/52) * (4/51)

Probability: [(1/52)*(3/51)] + [(12/52)*(4/51)] = 51/(52*51) = 1/52

now the other way:

Case 1: non-king spade and king
(12/52) * (4/51)

Case 2: king of spade and a king
(1/52)*(3/51)

Probability = 1/52

Hence, the total probability = 1/52 + 1/52 = 1/26.

Hope this helps:

Last edited by sanjayr84; 01-08-2007 at 04:11 PM. Reason: missed out another case
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medulla medulla is offline
It's time....
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Re: Probability question - 05-08-2007, 08:22 PM

Quote:
Originally Posted by vyomb View Post
hey medulla wats the ans for the ques..
Hi..

Sorry people..
My PC was not working.. So could not visit the forum..
The answer is 1/26...
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crack_this_time crack_this_time is offline
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Re: Probability question - 07-09-2007, 08:55 PM

A customer has to select a fruit basket of 20 fruits. Fruits can be mango, bannana or apple. How many ways can he select fruits?

(there can be zero fruits of one kind, key is M+B+A=20)

know this is kinda a easy, pls explain the method.

ciao
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Re: Probability question - 09-09-2007, 08:16 PM

Quote:
Originally Posted by sanjayr84 View Post
case 1: king of spade and a king
(1/52)*(3/51)

Case 2: non-king spade and king
(12/52) * (4/51)

Probability: [(1/52)*(3/51)] + [(12/52)*(4/51)] = 51/(52*51) = 1/52

now the other way:

Case 1: non-king spade and king
(12/52) * (4/51)

Case 2: king of spade and a king
(1/52)*(3/51)

Probability = 1/52

Hence, the total probability = 1/52 + 1/52 = 1/26.

Hope this helps:

i dont exactly get this solution. Why do we have to consider "now the other way:" ??
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Re: Probability question - 20-09-2007, 05:22 PM

Quote:
Originally Posted by crack_this_time View Post
A customer has to select a fruit basket of 20 fruits. Fruits can be mango, bannana or apple. How many ways can he select fruits?

(there can be zero fruits of one kind, key is M+B+A=20)

know this is kinda a easy, pls explain the method.

ciao

We know that non negative integral solution of the equation:
x1 + x2 + x3+ ....xr = N
are (N+r-1)C r-1

Answer in this case is 22C2 = 231.
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Re: Probability question - 20-09-2007, 10:06 PM

Hi friends,

Kindly help me solving these questions.

Thanks & regards,
niti


Q2)Seven blank spaces, which are numbered from 1 to 7,are aranged in a row in ascending order from left tp right.Two A's,three B's ,two C's are to b filled inthese seven blank spaces such that the same alphbet never occupy consecutive blank spaces.In how many ways can this be done if the B is always filled in the first blank space?
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crack_this_time crack_this_time is offline
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Re: Probability question - 25-09-2007, 09:17 PM

How many 5 digit numbers can be formed such that it has the following properties:
a)It has at least one zero and atmost three zeros.
b)The non zero digits are not repeated.

we have to find out for 3 cases
Case1: exactly 1 zero
Case2: exactly 2 zero
Case3: exactly 3 zero

what i dont seem to understand is the way the total number of such numbers is calculated.

take Case1:
step1) the first number (ten thousand's position) of the five digits can be placed in 9 ways (any number from 1 to 9)
step2) the second number(thousand's position) can be any number from 0 to 9 (except the one that has been putin the step1.) : total 9 ways
step3) the third number (hundreds position) 8 ways

step4) 7 ways
step 5) 6 ways

but where are we considering the 1 zero thats there in the number.

i m confused :(
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esh.nil esh.nil is offline
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Re: Probability question - 26-09-2007, 02:07 PM

Quote:
Originally Posted by niits View Post
Hi friends,

Kindly help me solving these questions.

Thanks & regards,
niti


Q2)Seven blank spaces, which are numbered from 1 to 7,are aranged in a row in ascending order from left tp right.Two A's,three B's ,two C's are to b filled inthese seven blank spaces such that the same alphbet never occupy consecutive blank spaces.In how many ways can this be done if the B is always filled in the first blank space?
@niti...Dude the answer to this qn is 20... the answer has been posted with the approach in the concepts-total fundas thread...


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esh.nil esh.nil is offline
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Re: Probability question - 26-09-2007, 02:21 PM

Quote:
Originally Posted by crack_this_time View Post
How many 5 digit numbers can be formed such that it has the following properties:
a)It has at least one zero and atmost three zeros.
b)The non zero digits are not repeated.
i m confused :(
The 3 cases are perfectly right,
1 zero:
we can select the 4 nos in 9*8*7*6 the zero can occupy any 4 places except the first digit, (4c1-- selecting 1 place from 4 available places)
so total no of ways = 9*8*7*6*4 = 12096

2 Zeros:
we can select the 3 nos in 9*8*7 now we have 5 places of which the first place cannot be occupied with zero. so eff 4 places and 2 zeros can be filled in those 2 places in 4c2 way
total no of ways = 9*8*7*6 = 3024

3 Zeros:
we can select the 2 nos in 9*8 ways and the 4 places can be filled with 3 zeros in 4c3 = 4 ways
total nos = 9*8*4 = 288

total nos = 12096+288+3024 = 15048
Is that correct???


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Re: Probability question - 27-10-2007, 09:31 PM

Quote:
Originally Posted by esh.nil View Post
The 3 cases are perfectly right,
1 zero:
we can select the 4 nos in 9*8*7*6 the zero can occupy any 4 places except the first digit, (4c1-- selecting 1 place from 4 available places)
so total no of ways = 9*8*7*6*4 = 12096

2 Zeros:
we can select the 3 nos in 9*8*7 now we have 5 places of which the first place cannot be occupied with zero. so eff 4 places and 2 zeros can be filled in those 2 places in 4c2 way
total no of ways = 9*8*7*6 = 3024

3 Zeros:
we can select the 2 nos in 9*8 ways and the 4 places can be filled with 3 zeros in 4c3 = 4 ways
total nos = 9*8*4 = 288

total nos = 12096+288+3024 = 15048
Is that correct???


you forgot to multiply by 4!, 3!, 2! in cases 1,2,3 resp
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