M40-21

Question

While dancing, Anna takes either one step forward or one step back, with a probability of \(\frac{1}{2}\) for each time. What is the probability that she ends up one step away from the original point at the end of 5 steps?

A. \(\frac{1}{10}\)
B. \(\frac{1}{5}\)
C. \(\frac{5}{16}\)
D. \(\frac{3}{8}\)
E. \(\frac{5}{8}\)

A.  \frac{1}{10} 
B.  \frac{1}{5} 
C.  \frac{5}{16} 
D.  \frac{3}{8} 
E.  \frac{5}{8}

Bunuel · Answer

Official Solution:

While dancing, Anna takes either one step forward or one step back, with a probability of \(\frac{1}{2}\) for each time. What is the probability that she ends up one step away from the original point at the end of 5 steps?

A. \(\frac{1}{10}\)
B. \(\frac{1}{5}\)
C. \(\frac{5}{16}\)
D. \(\frac{3}{8}\)
E. \(\frac{5}{8}\)

A.  \frac{1}{10} 
B.  \frac{1}{5} 
C.  \frac{5}{16} 
D.  \frac{3}{8} 
E.  \frac{5}{8}

To end up one step away from the starting point, Anna must take either three steps forward and two back, in any order, or three steps back and two forward, in any order.
 
The probability of the first scenario, three steps forward and two back, is  \frac{1}{2^5}*\frac{5!}{3!2!} = \frac{5}{16} . We multiply by  \frac{5!}{3!2!}  to account for all orders of the FFFBB sequence.
 
The probability of the second scenario, three steps back and two forward, is also  \frac{1}{2^5}*\frac{5!}{3!2!} . Again, we multiply by  \frac{5!}{3!2!}  to account for all orders of the BBBFF sequence.
 
Therefore, the total probability is  \frac{5}{16}+\frac{5}{16}=\frac{5}{8} .

Answer: E

phmahi1997 · Answer

I did not quite understand the solution. Why do we have to count 5!/(3!*2!)?

Bunuel · Answer

We use 5!/(3!*2!) because it counts all distinct ways to arrange 3 forward steps (F) and 2 backward steps (B) in 5 steps. For example:

If the steps are F, F, F, B, B, possible arrangements include:

FFFBB
FFBFB
BFFFB
FBFFB
...

In total, there are 10 such arrangements, which is why we calculate it as 5!/(3!*2!) = 10.

phmahi1997 · Answer

I like the solution - it’s helpful.

bumbleBee3445 · Answer

To get one step away from the origin possible scenarios are 1) 3 steps forward and 2 backwards or. 2) 3 steps backward and 2 forward.

The probability for taking 5 steps is (1/2)^5.
 no of different ways possible for  
1) 3 steps forward and 2 backwards FFFBB is 5!/3!2! (Reason we don't want duplicate orders of FFFBB hence dividing by that count)
 2) 3 steps backward and 2 forward BBBFF is 5!/3!2! (Reason we don't want duplicate orders of BBBFF hence dividing by that count)

Now actual prob that she ends up one step away from the original point at the end of 5 steps is = (1/2)^5 *( 5!/3!2! + 5!/3!2!)=1/32* 20= 20/32 = 5/8

Ans: E) 5/8

bumbleBee3445 · Answer

Here we are counting possible scenarios for arrangements FFFBB and BBBFF.

If these steps had been unique (like f1f2f3b1b2) then the possible arrangements would be 5! for FFFBB.
but here we have 3 same forward steps and 2 same backward steps, so we are overcounting by possible arrangements of FFF and BB, which are 3! and 2! respectively.

Hence we need to divide 5! by that number.

Hope this helps to clear your doubts.

getsanjeev · Answer

I like the solution - it’s helpful.

M40-21

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards