Quote:
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 + f
irst card non-heart and second card heart + first card a heart..
= 39c2+
39.13+13
favourable case = f
irst 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..