M05-17

Official Solution:

Kate and David each start with $10. They flip a coin 5 times. For each flip that results in heads, Kate gives David $1, and for each flip that results in tails, David gives Kate $1. After flipping the coin 5 times, what is the probability that Kate has more than $10 but less than $15?

A. \(\frac{5}{16}\)
B. \(\frac{15}{32}\)
C. \(\frac{1}{2}\)
D. \(\frac{21}{32}\)
E. \(\frac{11}{16}\)


After 5 flips, for Kate to have more than her initial sum of $10 and less than $15, she must win 3 or 4 times (if she wins 2 or fewer times, she will have less than $10, and if she wins 5 times, she will have $15).

So, the question becomes "What is the probability of getting 3 tails and 2 heads (TTTHH) or 4 tails and 1 head (TTTTH) in 5 flips?".

Therefore, \(P(TTTHH \ or \ TTTTH) = P(TTTHH) + P(TTTTH) = \frac{5!}{3!2!}*(\frac{1}{2})^5 + \frac{5!}{4!1!}*(\frac{1}{2})^5 = \frac{10}{32} + \frac{5}{32} = \frac{15}{32}\). We multiply by \(\frac{5!}{3!2!}\) for the first case and \(\frac{5!}{4!1!}\) for the second case because TTTHH and TTTTH can occur in that number of ways (for instance, the combination TTTTH can occur in \(\frac{5!}{4!1!}=5\) different ways: TTTTH, TTTHT, TTHTT, THTTT, and HTTTT. ).


Answer: B
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Bunuel, my approach to this problem was as follows:

There is an equal probability that Kate or David will end up with more money by the end (and note there is no situation where Kate and David both end up with $10). Therefore, we can subtract the probability that Kate wins all 5 flips from 1/2 to get the probability:

combinations = 2^5 = 32
Probability of winning all 5 flips = 1/32
Probability Kate ends up with $11-$14 = 1/2 - 1/32 = 15/32

If this game had been flipping an even number of coins, you'd have to do 1/2 - 1/64 - (probability of same number heads and tails).

Cheers,

I couldn't wrap my mind around how to apply combinatorics to this.

Instead, I figured that K and D both have 50% chance on each flip, therefore K success on the first 4 = 2 win: 2 loss. Probability of winning the tiebreaker (5th flip) is 50%

therefore she has 50% chance of winning 3/5 flips.

Lastly, K cannot win all 5 or she goes over the max. Prob 5/5 = (1/2)^5 = 1/32

Prob $0<K<$5 = 1/2 - 1/32 = 15/32

Dear Bunuel!
Please explain why the total number of combinations = 25^5 ?
I don't understand. Tks

They flip a coin 5 times. Each time it's either heads or tails, so 2 outcomes for each flip, there are 5 flips, so the total number = 2*2*2*2*2 = 2^5.

Hope it's clear.
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Kate looses $1 every time Head is flipped
Kate gains $1 every time Tail is flipped
Total combinations of coin flipping 5 times are 32 ways
Kate
H H H H H : Kate has $5 at end of 5 coin flips (Kate lost $5 for 5 head flip)
T T T T H : Kate has $13 (she gains $4 for 4 tails but looses $1 for 1 heads)
T T T H H : Kate has $12 (she gains $3 for 3 tails but looses $2 for 2 heads)
T T H H H : Kate has $9 (she gains $2 for 2 tails but looses $3 for 3 heads)
T T H H H : Kate has $9 (she gains $2 for 2 tails but looses $3 for 3 heads)
.
.
.
.
T T T T T : Kate has $15 (she gains $1 for 5 tails)

We are asked what is probability that Kate has more than $10 but less than $15

This is possible if there are atlest 3 or 4 Tails are flipped in 5 total coin flips
Number of ways 3 Tails flipped are 10
Number of ways 4 Tails flipped are 5

Probability = # of favourable outcomes / Total outcomes
= (10+5)/32

Kate wins denoted by K (say) and David's win denoted by D
So it can be the permutation of these two cases -
1. KKKDD
2. KKKKD

Now the first case is = \(\frac{5!}{3!2!} = 10\)
And the second case = \(\frac{5!}{4!} = 5\)
Total occurrence = \(2^5 = 32\)

Therefore, probability = \(\frac{10+5}{32} = \frac{15}{32}\)

Answer: C

Kate and David each have $10. Together they flip a coin 5 times. Every time the coin lands on heads, Kate gives David $1. Every time the coin lands on tails, David gives Kate $1. After the coin is flipped 5 times, what is the probability that Kate has more than $10 but less than $15?

Given:
Initially Kate has $10 and David has $10
Heads: Kate to david $1
Tails: David to Kate $1
Coin flipped 5 times

Find: Probability $10<Kate<$15

Case 1) Let suppose kate wins 1 game and David wins 4 game: KDDDD : After 5 toss, kate will have 7 and david will have 13: This case is not possible.
Case 2) Let suppose kate wins 2 game and David wins 3 game: KKDDD : After 5 toss, kate will have 9 and david will have 11: This case is not possible.
Case 3) Let suppose kate wins 3 game and David wins 2 game: KKKDD : After 5 toss, kate will have 11 and david will have 9: This case is possible. No of possible case : 5!/3!*2!= 10
Case 4) Let suppose kate wins 4 game and David wins 1 game: KKKKD : After 5 toss, kate will have 13 and david will have 7: This case is possible. No of possible case : 5!/4!=5
Case 5) Let suppose kate wins 4 game and David wins 1 game: KKKKK : After 5 toss, kate will have 15 and david will have 7: This case is not possible.

Total outcome: 32

Probability : 10+5/32=15/32

A. 5/16
B. 15/32
C. 1/2
D. 21/32
E. 11/16

It can also be solved without using P&C

So Kate and David have equal chances of winning. But the question has a restriction that Kate cannot win all 5 times. So the probability would be less than 1/2

Now comparing Answer choices we see

A. 5/16 >>too less (since the probability we are trying to find is eliminating only one possibility out of 5>>)
B. 15/32 >> Keep it
C. 1/2 >> cannot be since we are eliminating a possibility
D. 21/32 >> Greater than 1/2
E. 11/18 >> Again Greater than 1/2

Hence Ans is B

Hi IanStewart Bunuel chetan2u :

Please help me to identify the value which I get is actually what?

Out of 5 times coin flipped : Kate can have $11 or $13 , more than 10$ and less than 15$.
The values that she can have are: 5$, 7$, 9$, 11$, 13$,15$ ( if Kate wins 0, 1,2,3,4,5 times respectively)

Favored cases = 2/6 ( total cases) = 1/3

In other words, what would be the question exactly, if the answer is 1/3.

please suggest.

You might imagine using the same approach to answer the question "what is the probability I will win the lottery?" You might then think there are two cases -- you win or you lose -- and that the probability is 1/2. But that's clearly not right. The issue is, the two cases in the lottery question are not equally likely. If we're going to solve a probability question by counting outcomes, those outcomes need to be equally likely to happen. In the question in this thread, Kate only gets $15 when the coin lands on Tails five times, which can only happen in one way (TTTTT). But she gets $11 in the end if the coin lands on Tails three times and on Heads twice, which is a much more likely outcome -- it can happen in ten ways (TTTHH, TTHTH, THTTH, HTTTH, etc).



I can quickly modify the question above so 1/3 becomes the right answer, but I think this question setup is very awkward (I'd never write a question like this personally!) :

David and Kate each have $10. David and Kate play a game where they flip a coin once, and if the coin lands on Tails, David will give Kate a randomly selected even number of dollars between $2 and $6 inclusive, and if it lands on Heads, Kate will give David a randomly selected even number of dollars between $2 and $6 inclusive. What is the probability Kate will have more than $10 but less than $15 if she plays this game once?

Then there are six equally likely outcomes (-$6, -$4, -$2, +$2, +$4, +$6), two of which produce the result the question asks for, so the answer is 2/6 = 1/3.

I think this is a high-quality question and I agree with the explanation.
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I think this is a high-quality question and I agree with explanation.

Easy if done by letters.

Total number of outcomes is 2*2*2*2*2 = 32

Outcomes where K's money is between 11 and 14 ( 11 < k < 14)
w=win L=loss

WWWLL + WWWWL

total number of letters/repetition of letters

5! /(3!*2!) + 5! /(4!*1!) = ?
10 + 5 = 15

Answer = 15/32

Can someone please help me understand what's wrong with my below approach:

Possible amounts with Kate after 5 coin flips are:
- 15 (if Kate wins all 5)
- 13 (if Kate wins 4 and loses 1)
- 11 (if Kate wins 3 and loses 2)
- 9 (if Kate wins 2 and loses 3)
- 7 (if Kate wins 1 and loses 4)
- 5 (if Kate loses all 5)

Now only 11 and 13 (2) are the relevant ones out of 6 possibilities. So, it can be 2/6 or 1/3. Right?

There are more than six possibilities...
Lets take 13 as an example (4 wins and 1 loss) there is actually 5 ways for this outcome to come true...you are saying in your approach that there is only one way for getting this result.
-LWWWW
-WLWWW
-WWLWW
-WWWLW
-WWWWL

Now lets take 11 (3 wins and 2 losses) for this there are actually 10 possible outcomes that you are discounting as one again
-LLWWW, WLLWW, WWLLW, WWWLL, LWLWW, LWWLW, LWWWL, WLWWL, WLWLW, WWLWL,

Your reasoning would be correct if all of your 6 options had the same number of ways to get it but it is not the case

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Why are we Subtracting \(\frac{1}{2}\) in this equation?

In a given scenario, Kate wins in half of the cases, and David wins in the other half. Out of the cases in which Kate wins (1/2), we want to exclude the instance where she wins all five times (1/2^5). Therefore, the probability we are looking for is 1/2 - 1/32.
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