Rakesh is standing 3 steps to the right of a Green mark and 2 steps to

Question

Rakesh is standing 3 steps to the right of a Green mark and 2 steps to the left of a Pink mark. He tosses a coin. If it comes up heads, he moves one step to the left. If it comes up tails he moves one step to the right. He keeps doing this until he reaches one of the two marks, and then he stops. At which mark does he stop?

(1) He stops after 21 coin tosses.
(2) He obtains three more heads than tails.

(1) He stops after 21 coin tosses.
(2) He obtains three more heads than tails.

Azedenkae · Answer

I would think D. Since the question asks which mark he stops on after having finished tossing coins, it should be very simple. Unless he tossed an even number of coins, he must have stopped on the green mark. This is because he could not have moved an odd number of steps to get on the pink mark, and vice versa for the green mark.

Thus, with (1), because he tossed the coin 21 times, he would have had to have moved all the way left onto the green mark. Similarly, with (2), because he tossed 3 more heads than tails, all tail tosses would have been offset by head tosses, and then there's three extra to get him onto the green mark. For (2), this could also mean he may have tossed 0 tails, and thus still have just enough heads to get on the green mark.

So yeah, D.

SonalSinha803 · Answer

I don't know how to solve it in two minutes. I took 4 minutes minimum to solve this.

I think it should be B.

Statement 1, gives two possibilities with the stated condition. He could reach green as well as pink mark in 21 times toss. Just try impartially and see because it is sort of impossible to explain in writing how.

Statement 2 is sufficient. Seeing greater number of heads indicates that he cannot move to the pink mark. Thus, anytime he would reach the green mark only.

Thus,imo B.

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Jkim56 · Answer

Statement 1, I believe that considering there is 1/2 chance for heads and a 1/2 chance for tails, one can infer that there is an equal number of heads or tails + 1 (heads or tails) which makes it impossible to know if the person is at the green or pink marker.

Statement 2, taking into consideration that the probability for heads is 1/2 and the probability of tails is 1/2, one can infer that there is an equal number of heads or tails PLUS 3 more heads which would place the person at the green marker.

amanvermagmat · Answer

So this is the situation:

Green _ _ Rakesh _ Pink

Green mark is 3 steps to left of Rakesh and Pink mark is two steps to right of Rakesh.

Lets see in how many moves Rakesh and reach Pink mark. He can reach pink in either 2 moves (two rights), or 4 moves (one right, one left, then two rights), 6 moves (one left, one right, one right, one left, two rights).. and so on.. We can observe that Rakesh will always  land at Pink in even number of moves .. And  his 'right moves' have to be two more than his 'left moves' .

Now lets look at Green. Rakesh can reach green in either 3 moves (3 lefts), or 5 moves (2 lefts, one right, 2 lefts), or 7 moves (2 lefts, 2 rights, 3 lefts in any order).. and so on.. We can observe that Rakesh will always  land at Green in odd number of moves .. And  his 'left moves' have to be three more than his 'right moves' .

Now lets look at the statements:

(1) 21 moves and stopped. Since he stops only on reaching either Pink or Green and since 21 is an odd number, he has stopped at Green mark. Sufficient.

(2) 3 more heads than tails, so three more lefts than rights.. again Green as explained above. Sufficient. 
We can also look at it this way: if no of tails is 'x', then no of heads is 'x+3' and total = x+x+3 = 2x+3, which is an odd number. So has to be green.

Hence  D answer

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