Two coins one with HEAD IN BOTH SIDES and the other coin HEAD IN ONE SIDE AND TAIL IN THE OTHER SIDE is in a box,a coin is taken at random and FOUND HEAD IN ONE SIDE .what is the probability that THE OTHER SIDE IS HEAD?

Please answer this questions guys :::::::

1.Two coins one with HEAD IN BOTH SIDES and the other coin HEAD IN ONE SIDE AND TAIL IN THE
OTHER SIDE is in a box,a coin is taken at random and FOUND HEAD IN ONE SIDE .what is the
probability that THE OTHER SIDE IS HEAD?

2.12 Blacksox and 12 Whitesox mixed in a box,a pair of sox is picked at a time,in which pick/
how many pick ,to get the right pair(black&black or white&white)?

3.16 litre can, 7 litre can,3 litre can,the customer has to be given 11 litres of milk using
all the three cans only explain?

THANKS A LOT !!!

Questions by itcoll   answers by itcoll

Showing Answers 1 - 75 of 112 Answers

We hav three cans 16 litres, 7 litres, and 3 litres right...We wanna give 11 litres of milk using these 3 cans....Take first 7 litres can and add the milk to the container...now the contaier has 7 litres of milk..remaining is 4 litres...take 16 litres can and fill the can with the milk...using 3 litres can, take the milk from 16 litres can 4 times which means 12 litres of milk..therefore the left out milk in the 16 litres can is 4 litres..add tat milk to the container..now the total milk in the container is 11 litres

1. the probability is 0.5 (1/2).
Since the coin can be any one of the 2.
therefore, total sample space = 2
favourable case =1.

2. 3 picks would ensure that we get atleast one pair.

3. take 7L with help of one can.
Then take 7L in that can. Pour 3L in the 3L can.
Give remaining milk in 7L can (4L) to customer.
7+4=11

mani_12345

  • Jun 23rd, 2008
 

here a coin is taken at random and head is found for selecting a coin from 2 coins the probability is 1/2
and a head is already found in 1 side and the probability of getting tail on other side is
1st coin-0
2nd coin -1
so total probability is 1/2(1+0)=1/2

mani_12345

  • Jun 23rd, 2008
 

picking up sox here 12 black and 12 white are there soo we have to make a black and white pair so
1st selectting black from 12 black sox 12c1
2nd selecting white from 12 white sox 12c1
cumulation total successful pickups-12ci*12c1=144

1) The probability of selecting one coin is 1/2(which is already completed)
now.. The probability of having head on other side is (1/2) since only one coin has head. Therefore the prob. that head on other side is 1/2(second value from above explanation)

2) selecting 2 black ones is (no of ways) : 12c2
selecting 2 white ones is :12c2
therefore, selecting a pair of particular color is 12c2+12c2 = 66+66 = 132
selecting 1 black & one white is : 12c1*12c1 == 144

Lets assume worst case (which is required)
since 144 is greater than 132. 
the first pick will be black & white, 2nd pick is also black & white ,3rd will be black or white, no of picks required : 3

3) use 7 lt can and pour milk in 16lt can, now take milk in 7lt can, take 3lt from that 7lt can, 7lt can will have 4 lts in it, pour that 4 lt in 16lt can, it has 7(previous)+4(later) in it...

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dh.agrawal

  • Jun 30th, 2008
 

Answer is 3/8.
probability to come head on face is 3/4. but the question is what is the probability that other side is head?
since there are 2 coins, in which 1 had both side head and 2nd have one head n one tell. so the probability for other side to come head is 3/4 * 1/2 = 3/8
 if i am wrong, please correct me!!!!

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First pour milk into 7 litre can.from 7 litre can pour milk to 3 litre can...the milk left over in 7 litre can is now 4 litres..now pour this 4 litres of milk to 16 litre can...fill the 7 litre can again and add it to the 16 litre can which gives us a total of 4+7=11 litres.

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ans.1-> probablity shud be half
ans.2->3 picks
ans.3->take 7 litre in 7litre can.take out 3 litre.put rest 4 litre in 16 litre can and add more 7 litre to it making 11 litre in 16 litre can..

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Well first there are 2 coins....

prob=no of favorable outcomes/sample space

probability of choosing one of the 2 coins is=1/2

if the first coin was chosen(H in both sides)probability of getting a head is =2/2=1

if the second coin was chosen prob of getting head is=1/2

hence total probability of getting a head =1/2*1 + 1/2*1/2=(prob of getting first coin*prob of getting heads)+(prob of getting second coin*prob of getting heads)=3/4 !!!

so many wrong answers before.

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Assuming that there are no markings on the cans to denote no of litres

-->First, use the 7 litre can to fill the 16 litre can with 7 litres of milk

-->Second, use the 7 litre can to fill the 3 litre can with 3 litres of milk

-->third ,Pour the 3 litres of milk from the 3 litre can into the 16 litre can(which already contains 7 litres poured during first step)

NOTE:1) now the 16 litre can has (7+3)=10 litres of milk
          2) the 7 litre can has 4 litres of milk left over from second step

-->fourth,now fill the 3 litre can with the rest of the milk(4 litrs) from 7 litre can 
    
     NOTE:Only one litre should be left out in the 7 litre can

-->Pour the 1 litre from 7 litre can into the 16 litre can ..Hence u have it (7+3+1)=11 !!

wanderer11

  • Aug 7th, 2008
 

As getting a head on one face of the coin we picked is a certainty...the probability that the other side is head will only be 1/2.

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vijum

  • Jan 29th, 2009
 

First fill the 3 litre can and pour that milk into 7 litre can,
Do it for two times now the 7 litre can contains 6 litres of milk,
Now again take milk in the 3 litre can and pour it into the 16 litre can for three times so the 16 litre can will be containing 9 litres of milk (3*3),
So now 16 litre can containing 9 litres milk and 7 litre can containing 6 litres of milk.
Now again take milk in 3 litre can and pour it in to the 7 litre can untill it is getting filled (already 7 litre can containing 6 litres of milk) so remaining milk in the three litre can will be only two litres

Pour it in to the 16 litre can which already having 9 litres of milk so 9+2=11 litres.

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Definitely it will be 1/2
As we have selected one coin and there is HEAD and now there are two possibilities that other side may be HEAD or TAIL means 2 cases and we required one out of these two
? ? so probability is ?= required case(1)/all possible case(2)
=1/2 ? ans

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As there is 11 litre milk ans can of 16,7,3
1) Take 7 litre milk in 7 litre can
2)
Now shift the 3 litre milk in 3 litre can
3) Now shift the remaining 4 litre milk in 16 litre can
4)
Shift the milk of 3 litre can into 7 litre can which is empty
5) Now fill it up to top means there will be 3+4=7 litre milk in this can
6)
Now shift this 7 litre can into 16 litre can ,,,already there was 4 litre milk so now it will be 4+7=11 litre milk in 16 litre can

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vasuki.v

  • Apr 22nd, 2009
 

1/2.

since when a coin with both sides head is taken the probability of getting head is 1/2.
when the other coin is taken the probability is 0.
hence or function 1/2+0=1/2

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Your answer would be 2/3 because in this we have two coins but the probability of being a head comes out from three ways as we are left with one coin and it may be a H-T coin or H-H coin so we are left with three position that is H H T so probability of getting a Head is 2/3.

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Total number of coins are two.
We took only one coin with one side as head.
But in that box we had one with two sides of head and other is one head and
tail.
There is only one coin having both side as head, so the answer is 1/2


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jkuppal26

  • Dec 3rd, 2009
 

Add milk 7 litre can
Put this into 16 litre can
Again add milk to 7 litre can
Put this into 16 litre can
Now 16 litre can has 14 litre milk
Add milk from this can to to 3 litre can
What is left is 11 litre

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1) probability 1/2

2) picks 3

3) Add 2x 7L to the 16 litre can and from that fill the 3L can
so that the remaining will be 11L... 2(7L)-3=11L   

Simple!

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subutheboss

  • Dec 16th, 2009
 

Take the 7litre can fill the container for remaining 4litre,
Fill the 7litre can then pour oil from 7litre can to 3litre can,
so remaining amt of oil in 7litre can is 4litre,
so now fill the remaining 4litre

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ramsrib

  • Dec 18th, 2009
 

We have 16, 7 and 3 liter cans, So first take 7 litre can and fill milk into it. after that from 7 litre take 3 litre of it using 3 litre can. pour the remaining 4 litre from 7 litre can to 16 litre can. Then add 7 litre to the 16 litre can. We get finally 11 litres.



7-3 =  4 liter ==> 16 liter can 
7 liter ==>  16 liter can (4+7 = 11 liters)


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anon1987

  • Jan 29th, 2010
 

the answer is 3/4

The probability of getting head of the coins is 1 & 1/2....

Gettng head on the real coin is done by conditional Prob...

P(H/H)=P(H)=1/2

Gettng head on the fake coin is done by conditional Prob...
P(H) = 1..


Getting the head prob .. = 1/2( real+fake)
                                     =1/2(1/2+1)
                                     =1/2(3/2)= 3/4 (ans)

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Ans:1;
Because there is two coins, one with two heads, so there is two chances in 1st coin and another coin has one side is head and another is tail so there is a chance of one head but the question is they takes a coin at random that means any one coin has taken in that they have one side is head. So we have chance in 1st coin which have both side heads then if you choose another there is no chance because they showed one side is head so there will be no head it will be tail on second coin, hence the total probability also 1.

The perfect answer is 1/2 and 0/1 (1 opportunity of being head/2 no of heads) and (0 opportunity heads/1 no of heads)

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prem_bombay

  • Aug 31st, 2010
 

First, Fill the 7 litre can with milk and then pour that in 16 litre can.
Again do the same process. So now 16 litre can contains 14 litre of milk.
Now take a 3 Litre can and fill it with a milk from 16 litre can. Hence now 16
litre can contains 14 - 3 = 11 litre of milk.


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Nasim2

  • Sep 1st, 2010
 

2.     n(s)=24c2=12*23


        n(E)=12c2+12c2=2*12c2=12.11

probability=p(a)=(12*11)/(12*23)=11/23

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murali45

  • Oct 1st, 2010
 

that means one head is already fixed,,,from remaining two coins we should choose 

1tail,2 heads
probability being a head is 2/3 

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