Probability question

[esh.nil]

Originally Posted by brigadier

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

@brigadier.. It shall be 4 only...
what i did was I formed all possible 4 digit nos and then appended a 0 like ABCD0, this 0 can take any 4 place i.e
ABC0D
AB0CD
A0BCD
hence multiplied with 4...
Same for the other 2 cases,
Pls revert if I am wrong..

[santhosh_kumard]

Guys i have a couple of questions. I derived an answer, but looking at the options i realised that i was nowhere near, so need the help of the guru;s over here. So here we go

(i) A race where 12 horses are running, the chance that
- Horse A will win is 1/6
- Horse B will win is 1/10
- Horse C will win is 1/8.
Assuming that a tie is not possible, find the chance that one of them will win ( A, B or C )
Options -> (a) 47/120 (b) 1/480 (c) 1/160 (d) 1/240

(ii) A & B pick up a card at random from a weel shuffled pack of cards one after the
other, replacing it every time till one of them gets a heart. If A begins the game , then
find the probability that the game ends with B
Options -> (a) 3/7 (b) 4/7 (c) 3/4 (d) 1/4

Please help fellow puys!

[esh.nil]

Originally Posted by santhosh_kumard

Guys i have a couple of questions. I derived an answer, but looking at the options i realised that i was nowhere near, so need the help of the guru;s over here. So here we go

(i) A race where 12 horses are running, the chance that
- Horse A will win is 1/6
- Horse B will win is 1/10
- Horse C will win is 1/8.
Assuming that a tie is not possible, find the chance that one of them will win ( A, B or C )
Options -> (a) 47/120 (b) 1/480 (c) 1/160 (d) 1/240

(ii) A & B pick up a card at random from a weel shuffled pack of cards one after the
other, replacing it every time till one of them gets a heart. If A begins the game , then
find the probability that the game ends with B
Options -> (a) 3/7 (b) 4/7 (c) 3/4 (d) 1/4

Please help fellow puys!

For the first one, I will choose option (a), bocs the individual probablity is 1/6,1/8,1/10, hence the prob that P(a),P(b),P(c) shld be def greater than 1/6
hence the option (a) alone satisfies the condition. I will post if the soln strikes.

For the second one, i feel they shld not replace the card.. If they are replacing the card, then the possibilities are infinte. consider this case,
they pick a non-heart card and then replace it and then keep picking the same card indefinitely, so the probability is inderterminate in that case, hence i feel the problem shld be considered assuming that there is no replacement of the card...
so assuming they dont replace the card,
the problem becomes find the probability of drawing 2 cards such that the second card is a heart..
total cases = drawing 2 non-heart cards + first card non-heart and second card heart + first card a heart..
= 39c2+39.13+13
favourable case = first card non-heart and second card heart = 39.13
probability = 39.13/(39c2+39.13+13)
but this is no where close to the soln.. I am not sure where I am going wrong.. or is my assumption absurd? pls help..

[STALWART]

Originally Posted by santhosh_kumard

Guys i have a couple of questions. I derived an answer, but looking at the options i realised that i was nowhere near, so need the help of the guru;s over here. So here we go

(i) A race where 12 horses are running, the chance that
- Horse A will win is 1/6
- Horse B will win is 1/10
- Horse C will win is 1/8.
Assuming that a tie is not possible, find the chance that one of them will win ( A, B or C )
Options -> (a) 47/120 (b) 1/480 (c) 1/160 (d) 1/240

(ii) A & B pick up a card at random from a weel shuffled pack of cards one after the
other, replacing it every time till one of them gets a heart. If A begins the game , then
find the probability that the game ends with B
Options -> (a) 3/7 (b) 4/7 (c) 3/4 (d) 1/4

Please help fellow puys!

( I )

See it is the case of MUTUALLY EXCLUSIVE and EXHAUSTIVE case ....
It is as good as saying that ..... Four horses are taking part in race namely .. A, B , C and D .

P( winning of A ) = 1/6
P( winning of B ) = 1/10
P( winning of C ) = 1/8
P( winning of D ) = 1- { P(A) + P(B) + P(C) } = 73/120

As in EXHAUSTIVE events ... Sum of all the probabilities is equal to 1.
So .. here if any one of them wins , the probability of winning of other is 0 as tie cudnt be there.

So ,

P ( A or B or C ) = ( 1/6 ) + ( 1/10 ) + ( 1/8 ) = 47/120
Or
Say P ( of NOT winning of D ) = 1 - P( winning of D ) = 1- ( 73/120 ) = 47/120


( II )

Here P ( Heart ) = 13/52 = 1/4 And
P ( NON-Heart ) = 39/52 = 3/4

Here A is starting the game and we have to find out that B wins the gaming picking up any Heart . Mind it ... Card is being placed in the pack everytime .. So it will not affect the Probabilities of picking up of Heart aor Non-heart in any draw.

P (B) = [P(NON-Heart byA)*P(Heart by B)] + [P(NON-Heart by A)*P(NON-Heart by B)*P(NON-Heart by A)*P(Heart by B)] + ..................

P(B) = (3/4)(1/4) + (3/4)(3/4)(3/4)(1/4) + ...............

It will form a GP ............. Solve it

==>> P(B) = 3/7

... Sayonara

[xcoolaryan]

Arun Sharma, LOD I - Pbm 43

Kamal & monica appeared for interview for 2 vacancies. Probability of Kamal's selection is 1/3 and that of Monica is 1/5. find probability if only one of them is selected?

a. 2/5 b. 1/5 c.5/9 d.2/3

[STALWART]

Originally Posted by xcoolaryan

Arun Sharma, LOD I - Pbm 43

Kamal & monica appeared for interview for 2 vacancies. Probability of Kamal's selection is 1/3 and that of Monica is 1/5. find probability if only one of them is selected?

a. 2/5 b. 1/5 c.5/9 d.2/3

P = (1/3)*(4/5) + (2/3)*(1/5)

" * " = AND
" + " = OR
==>> p = 2/5 !!

... Sayonara

[shivgudia]

The probabilities that a student pass in Mathematics, Physics and Chemistry are m, p, c respectively.
Of these subjects the student has a 75 % chance of passing in at least one, a 50% chance of
passing in at least two and 40 % chance of passing in exactly two. Which of the following relations
are true ?
a. p + m + c =19/20
b. p + m + c =27/20
c. pmc =1/10
d. pmc =1/4

[pallaviagrawal]

Originally Posted by shivgudia

The probabilities that a student pass in Mathematics, Physics and Chemistry are m, p, c respectively.
Of these subjects the student has a 75 % chance of passing in at least one, a 50% chance of
passing in at least two and 40 % chance of passing in exactly two. Which of the following relations
are true ?
a. p + m + c =19/20
b. p + m + c =27/20
c. pmc =1/10
d. pmc =1/4

from the given conditions we have..
m+p+c+mp+pc+mc+mpc(probability of passing in at least one)=.75
mp+pc+mc+mpc(at least 2)=.5
and mp+pc+mc(xactly 2)=.4

solving the 3 equations we get mpc=.10 or 1/10
c. is d right ans

[krishna13]

One box contains 8 cubes, and 6 spheres. A second box contains 6 cubes and 12 spheres. One object is transferred from the first box to the second box.
Probability that an object selected from second box to be a cube is?

Ans: 46/133

Coud anyone please help me with this question?

[the_hate]

Originally Posted by krishna13

One box contains 8 cubes, and 6 spheres. A second box contains 6 cubes and 12 spheres. One object is transferred from the first box to the second box.
Probability that an object selected from second box to be a cube is?

Ans: 46/133

Coud anyone please help me with this question?

case 1- obj transferred is cube.
prob = (8/14)*(7/19) = 28/133

OR

case 2-onj transferred is sphere.
prob = (6/14)*(6/19)=18/133

since or case....we'll add the prob....
total prob = 46/133