The figure above shows a zebra crossing made of alternate black and

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The figure above shows a zebra crossing made of alternate black and white stripes. A frog sits on the middle black stripe and on each jump, it can only move from one stripe to a stripe next to it, right or left. What is the probability that after 25 jumps the frog returns to the middle black stripe?

A. 0
B. 1/25
C. 2/25
D. 13/25
E. 1

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A

kunalcvrce · Answer

GST Week 1 Day 1 Exampal Question 1 
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The figure above shows a zebra crossing made of alternate black and white stripes. A frog sits on the middle black stripe and on each jump, it can only move from one stripe to a stripe next to it, right or left. What is the probability that after 25 jumps the frog returns to the middle black stripe?

A. 0
B. 1/25
C. 2/25
D. 13/25
E. 1

For moving forward and backward each trip will be a factor of 2.

but total 25 is odd .So probability that after 25 jumps the frog returns to the middle black is 0

option A

subsauce · Answer

The answer is A. 0 percent chance.
If you tried to figure this out mathematically it may give you a headache. Instead using logic you will find that it is impossible to land on any black stripe with 0 moves left.

Assuming | = black strip, _ = white stripe , and x= Frog, and that the chart looks like this |_|_|_X_|_|_|

On the first move, the frog will land a white stripe (or _) with an even amount of moves left: |_|_|_|X|_|_| or |_|_|X|_|_|_|

Considering 0 is even, that means with 1 move left you NEED to land on a white stripe which cannot happen.

To further explain this lets make this problem easier: Lets change the strips to |_X_| and only allow 5 moves.

5 moves left: |_X_| 
4 moves left: |_|X| or |X|_|
3 moves left: |_X_| or X_|_| or |_|_X 
2 moves left: |_|X| or |X|_|
1 moves left: |_X_| or X_|_| or |_|_X 
0 moves left: |_|X| or |X|_|

US09 · Answer

Every to and fro move of frog will be a multiple of 2. To come back to the initial positions, the number of jumps required will therefore always be in multiples of two. For example, 2, 4, 6, 8..2*n (n=no of jumps). Hence, as 25 is odd, it will always result in frog moving away from the previous positions of jump.

Therefore, A=0 is the probability.

Therefore, A=0 is the probability.

seed · Answer

Scenario 1: If the frog jumps to the right black strip and comes back then it takes 2 jumps.
Scenario 2: If the frog jumps to second right black strip and comes back then it takes 4 jumps.
Scenario 3: If the frog jumps to third right black strip and comes back then it takes 6 jumps.

Please note that above scenarios are valid for black strips to the left of frog.

We see here that for jumping and coming back to the center the frog takes even jumps (2,4,6) so, for given odd jumps .i.e. 25 the frog will never come back to the center black strip.

The probability is 0. (A).

souvonik2k · Answer

Let us number the black and white stripes as follows, with the number 0 representing the middle black stripe, and 1, 3, 5 representing black stripes. Similarly, 0,2,4,6 represents the black stripes.

5 4 3 2 1 0 1 2 3 4 5 6

Observe that 25 is an odd number, and hence the frog cannot reach the center black stripe even if it makes 25 jumps.
Hence the probability must be 0.
Answer A.

Digvijay101 · Answer

As the number of jumps is odd-25, there is 0% chance of the frog returning to the starting position

pawanpoolla · Answer

The answer is  Option A  :  0

Here is the solution: 
For questions like these which seem impossible to solve it is usually trying to count is the way to go - these usually lead to a pattern.
Lets see how the frog may return to the middle stripe:
1. Say it jumps 1 left and then jumps back to the middle stripe - that's 2 jumps. Repeating the same it's 4, then 6.... so essentially multiples of 2. So this way it will need either 24 or 26 jumps to return to the middle stripe.

2. Next consider that it jumps 2 left and then back 2 to the middle stripe - that's 4 jumps. Repeating the same it's 8 jumps, then 12. Again multiples of 4 - So either 24 or 28 jumps this way

Similarly try by adding one extra jump to the left - you will see a pattern in the number of jumps required to return - 2, 4,6, 8, 10, 12....etc. Note that since the frog is at the center - moving left or right will be symmetrical.

Finally, there may be cases that the frog not repeat the same pattern i.e. it may do 2 left and 4 right and return to the middle stripe i.e. in any combination of the above mentioned jumps. However, if you recall - sum of two even numbers is even. So any combination will result in an even number - meaning that the 25th jump will always be away from the middle stripe.

Hence, 0 cases where the frog returns to the middle stripe after the 25 jumps. So the probability is 0.

rahul16singh28 · Answer

Every Jump to the Middle Black Stripe is a multiple of 2. Hence, Probability = 0.

gmatexam439 · Answer

Answer must be 0 because frog will be on black strip every even number of jump. Since 25 is odd, the frog will be on white.

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mazeltov · Answer

A. 0 percent

Assuming that the middle black stripe is 0, and each right jump or left jump equals +1 or -1 respectively. If you want 0 as the final result, you need as many positive value as negative value, which means an equal number of left and right jump made. Consequently, the total number of left and right jump will be a multiple of 2, which can never be an odd number like 25.

gmatasp12 · Answer

The probability is 0.
The zebra needs steps in the multiple of 2 to come back to the middle black strip and 25 is an odd no and not a multiple of 2.

Posted from my mobile device

souvik101990 · Answer

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The figure above shows a zebra crossing made of alternate black and

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