M05-17

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.

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.

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.

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.