@Shray14 said:A shopkeeper mixed coffee and chikori powder in the ratio of 5:3 in a jar of 40 Kgs. How much chikori powder should he mix to these 40 kgs. To make the ratio 1:1?
[(3/8)40]+x=(5/8)*40, x=10
@Shray14 said:A music shop sells cassettes of English music, Hindi film songs and Telegu film songs at a ratio of 2:4:5. If in the month of June the shop sold 452 cassettes of Hindi film songs, the number of cassettes of Telegu film songs is??
No of telegu film songs= 452*(5/4)=565
Dagny's purse contains 4 coins which can be 1Rs coins or 5Rs coins. 2 coins are drawn and are both found to be 5Rs coins. If the coins are replaced and an another coin is drawn, what is the probability that it is a 1Rs coin ?
@hiteshkhurana82 said:Dagny's purse contains 4 coins which can be 1Rs coins or 5Rs coins. 2 coins are drawn and are both found to be 5Rs coins. If the coins are replaced and an another coin is drawn, what is the probability that it is a 1Rs coin ?
Is the answer 3/4 ??
@jnirmal07 said:@soumitrabengeri15 kg of chikori powder should be added@soumitrabengeri
Please check ur answer again..10 kg will have to be added
Answer cannot exceed 1/2 since 2 out of 4 coins are rs5
since question has no info on the other 2 coins, probability cant be calculated.
there can be 2 coins of each
or
there can be 3 coins of rs5 and 1 coin of rs1
or
there can be 4 coins of rs5
@Maxray2 said:Answer cannot exceed 1/2 since 2 out of 4 coins are rs5
Number of 1 rupee coin can be (2 out of 4) OR (1 out of 4) or (0 out of 4)...nahi kya ???
@2ndApr2011 said:Number of 1 rupee coin can be (2 out of 4) OR (1 out of 4) or (0 out of 4)...nahi kya ???
You are correct..
probabily can be 2/4=1/2
or
1/4
or
0/4 (all are rs5 coins)
probabily can be 2/4=1/2
or
1/4
or
0/4 (all are rs5 coins)
@Maxray2 said:Answer cannot exceed 1/2 since 2 out of 4 coins are rs5
@gautam22 said:1/4??
ans = 1/8
In the image below: - AB and AD are tangent to the circle - BC and AD are parallel
What is the length of AC?
What is the length of AC?
A detective plays Russian roulette with a criminal. He has a revolver with a cylinder which can hold 6 bullets, but only one of the chambers is loaded. The detective starts the game and spins the chamber, then points the muzzle at himself and pulls the trigger, then the criminal does the same. However once the particular chamber has been accessed, it cannot be accessed again. So if detective pulls trigger on chamber 1, when it comes to the criminal, the criminal can randomly select only one of the 5 remaining chambers. The game ends when one of them dies. Who has a higher probability of winning, and by what percentage is the winner's probability of winning, greater than the loser's probability of winning?
Post approach please!
@hiteshkhurana82 said:Dagny's purse contains 4 coins which can be 1Rs coins or 5Rs coins. 2 coins are drawn and are both found to be 5Rs coins. If the coins are replaced and an another coin is drawn, what is the probability that it is a 1Rs coin ?
A1=5555, A2=5551, A3=5511; E=Select 2 5rs coins out of 4; P(M)=probablity of selecting 1rs coin
P(EA1)=1, P(EA2)=3C2/4C2=1/2, P(EA3)=2C2/4C2=1/6
P(A1)=1/(1+1/2+1/6)
P(A2)=1/2/(1+1/2+1/6)
P(A3)=1/6/(1+1/2+1/6)
P(THAT A COIN DRAWN IS OS 1rs)=P(A1)*P(M/A1)+P(A2)*P(M/A2)+P(A3)*P(M/A3)
=3/5*0+(3/10)*1/4+(1/10)*1/2=1/8
P(EA1)=1, P(EA2)=3C2/4C2=1/2, P(EA3)=2C2/4C2=1/6
P(A1)=1/(1+1/2+1/6)
P(A2)=1/2/(1+1/2+1/6)
P(A3)=1/6/(1+1/2+1/6)
P(THAT A COIN DRAWN IS OS 1rs)=P(A1)*P(M/A1)+P(A2)*P(M/A2)+P(A3)*P(M/A3)
=3/5*0+(3/10)*1/4+(1/10)*1/2=1/8
@anantn said:A detective plays Russian roulette with a criminal. He has a revolver with a cylinder which can hold 6 bullets, but only one of the chambers is loaded. The detective starts the game and spins the chamber, then points the muzzle at himself and pulls the trigger, then the criminal does the same. However once the particular chamber has been accessed, it cannot be accessed again. So if detective pulls trigger on chamber 1, when it comes to the criminal, the criminal can randomly select only one of the 5 remaining chambers. The game ends when one of them dies. Who has a higher probability of winning, and by what percentage is the winner's probability of winning, greater than the loser's probability of winning?Post approach please!
Let the probablilty of detective dying be D and for criminal dying be C
hence D= 1/6 (when he s shot in the 1st chance)
+ 5/6(when he s not shot in the 1st chance) * 4/5(criminal is not shot in the 1st chance) * 1/4(when he s shot in his 2nd chance)
+ 5/6 * 4/5 * 3/4 * 2/3 * 1/2 = 1/2
Similarly calculate C and it comes to be 1/2
Hope i'm correct
??@anantn said:A detective plays Russian roulette with a criminal. He has a revolver with a cylinder which can hold 6 bullets, but only one of the chambers is loaded. The detective starts the game and spins the chamber, then points the muzzle at himself and pulls the trigger, then the criminal does the same. However once the particular chamber has been accessed, it cannot be accessed again. So if detective pulls trigger on chamber 1, when it comes to the criminal, the criminal can randomly select only one of the 5 remaining chambers. The game ends when one of them dies. Who has a higher probability of winning, and by what percentage is the winner's probability of winning, greater than the loser's probability of winning?Post approach please!
p(detective wins)=5/6*1/5+5/6*4/5*3/4*1/3+5/6*4/5*3/4*2/3*1/2*1=1/6+1/6+1/6=1/2
p(criminal winning)=1/6+5/6*4/5*1/4+5/6*4/5*3/4*2/3*1/2=1/6+1/6+1/6=1/2
same for both
p(criminal winning)=1/6+5/6*4/5*1/4+5/6*4/5*3/4*2/3*1/2=1/6+1/6+1/6=1/2
same for both
krum and anurag1701 : spot on :)
@anantn said:A detective plays Russian roulette with a criminal. He has a revolver with a cylinder which can hold 6 bullets, but only one of the chambers is loaded. The detective starts the game and spins the chamber, then points the muzzle at himself and pulls the trigger, then the criminal does the same. However once the particular chamber has been accessed, it cannot be accessed again. So if detective pulls trigger on chamber 1, when it comes to the criminal, the criminal can randomly select only one of the 5 remaining chambers. The game ends when one of them dies. Who has a higher probability of winning, and by what percentage is the winner's probability of winning, greater than the loser's probability of winning?
Post approach please!
1/2 aa rahi hai for both
will share approach if correct
@Shray14 said:A shopkeeper mixed coffee and chikori powder in the ratio of 5:3 in a jar of 40 Kgs. How much chikori powder should he mix to these 40 kgs. To make the ratio 1:1?
25 kg of mixed coffee..and 15kg of chikori
so 10 kg of chikori will be added to make ratio 1:1