Algebra - probability

62

Lokesh Verma ·2009-02-20 00:23:39

case 1: only 1 even number

caes 2: no even number

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Ð”Ð²Ò¥Ñ—ÑuÏ now in medical c ·2009-02-21 00:34:35

[(3/4)^N+(3^N-1*N)/4^N]*[1-(3/4)^N] + [1- (3/4)^N+(3^N-1*N)/4^N]*(3/4)^N

[2]

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62

Lokesh Verma ·2009-02-20 04:29:43

Abhirup I guess I will save the answer for the question of the day? ;)

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1

ankit mahapatra ·2009-02-20 04:15:43

Even i was in doubt therefore i asked if the answer was 2/3.
I was not satisfied by my answer but was not able to make out the flaw in my answer. Thanks for pointing it out.

I am giving two different answers in two different sites!! he he

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62

Lokesh Verma ·2009-02-20 04:10:11

no ankit that solution is flawed...

Because take another example

The probability of getting 1 on throwing a dice..

There are 2 cases.

1st case : getting a 1 ...
2nd case: not getting a 1

so the probability is 1/2 which is because of the two cases only one is satisfying

This is obviously wrong because the probability is 1/6 and we know it very well

You are making the same mistake here :)

I hope you understand :)

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13

Ð”Ð²Ò¥Ñ—ÑuÏ now in medical c ·2009-02-20 04:09:10

yup...it must be 3/4

bhaiyaa...watz the ans of ur 2nd one???

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1

big looser ......... ·2009-02-20 04:08:33

tution notes

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62

Lokesh Verma ·2009-02-20 04:06:39

answer is not correct aman.

I have no doubt.

Tell me the source.

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1

ankit mahapatra ·2009-02-20 04:06:22

we have to take three cases :

O-odd
E-even

OO,OE,EE ( we can take either OE or EO because both are same cases because we should not be bothered about sequence of numbers)

total sample space- 3
favorable outcomes - 2 .
therefore
P(product is even)= 2/3

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1

big looser ......... ·2009-02-20 04:02:04

answer is 2/3

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1

ankit mahapatra ·2009-02-20 03:48:10

can the probability be 2/3 for 1st question?

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62

Lokesh Verma ·2009-02-20 00:40:24

for N=1, your probabiliyt is zero ;)

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62

Lokesh Verma ·2009-02-20 00:40:00

@abhirup...

what if the even number is a multiple of 4?

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13

Ð”Ð²Ò¥Ñ—ÑuÏ now in medical c ·2009-02-20 00:31:01

1-[(N+1)/2^N]

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Ð”Ð²Ò¥Ñ—ÑuÏ now in medical c ·2009-02-19 23:32:48

3/4

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13

Ð”Ð²Ò¥Ñ—ÑuÏ now in medical c ·2009-02-20 00:21:31

[2]

[12]

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62

Lokesh Verma ·2009-02-20 00:18:19

no abhirup

there will be 2 cases..

think again...

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13

Ð”Ð²Ò¥Ñ—ÑuÏ now in medical c ·2009-02-20 00:15:41

What is the probability that the product is a multiple of 4?
2^N-2/2^N[7]

so...1/4

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13

Ð”Ð²Ò¥Ñ—ÑuÏ now in medical c ·2009-02-19 23:55:06

see...choosing an even or and odd is equiprobable...na...there is 50% chance of picking an odd no.

the product is odd only when both are odd

the the probability of getting odd product is 1/2*1/2=1/4

the probability of getting even product is 3/4

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1

big looser ......... ·2009-02-19 23:50:58

both

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13

Ð”Ð²Ò¥Ñ—ÑuÏ now in medical c ·2009-02-19 23:48:50

which one is not clear to you??....ur Q??or bhaiya's??

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1

big looser ......... ·2009-02-19 23:47:30

clear nai hua

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13

Ð”Ð²Ò¥Ñ—ÑuÏ now in medical c ·2009-02-19 23:40:19

if all N are odd only then the product is odd

so ans of the first one 1-(1/2^N)

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1

chemistry organic ·2009-02-19 23:39:36

OR IT SHOULD B 2/3.......TOTAL 3 PAIRS
1.ODD-ODD
2.EVEN-EVEN
3.ODD-EVEN

......WILL THERE B ONE MORE CASE "EVEN-ODD".....CONFUSED....

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62

Lokesh Verma ·2009-02-19 23:36:31

din see ur solution abhirup :)

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62

Lokesh Verma ·2009-02-19 23:36:14

good work abhirup

@ aman:

This can be done like this

only way a product will be odd is chosing both numbers as odd.

that is of probability 1/2 x 1/2 = 1/4

so actual prob of the product being even is 1-1/4

Now Try this question..

N integers are chosen randomly and multiplied

What is the probability that the product is even.
What is the probability that the product is a multiple of 4?

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13

Ð”Ð²Ò¥Ñ—ÑuÏ now in medical c ·2009-02-19 23:35:24

the product is odd when both are odd.....probability of this is 1/2*1/2=1/4

so...required probability 1-1/4=3/4

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