Probability Chip Tossing Problem with Different Chip Tossing Counts Mandy and Chuck are flipping are flipping chips. The chip has two sides: head and tails. Each round Mandy flips 3 chips and Chuck flips 2 chips. Each round, they might throw an equal number of heads. In the round they first flip different numbers of heads, what is the probability Mandy threw more heads than Chuck?
My work:
I started off by listing the possible outcomes.
Mandy: HHH, HHT, HTH, THH, TTH, THT, HTT, TTT
Chuck: HH, HT, TH, TT
There are 8*4 = 32 possible outcomes in terms if rounds
I listed possible outcomes where Mandy threw more heads than Chuck.
1.HHH, HH
2.HHH, HT
3.HHH, TH
4.HHH, TT
5.HHT, HT
6.HHT, TH
7.HHT, TT
8.HTH, HT
9.HTH, TH
10. HTH, TT
11. THH, HT
12. THH, TH
13. THH, TT
14. TTH, TT
15. HTT, TT
I got 15 different possibilities. Does this mean the answer is 15/32.
Thank you very much
I'm not really sure how to approach this problem?
 A: Let $A$ = the event "They throw different numbers of heads"
Let $B$ = the event "Mandy threw more heads than Chuck"
$P(B)$ is what you computed, $15/32$.
You want $P(B|A) = \frac{P(B \cap A)}{P(A)}$
Let $M$ = the number of heads Mandy gets
Let $C$ = the number of heads Chuck gets
$P(M=C) = P(M=C=0) + P(M=C=1) + P(M=C=2)$
Use binomials in general to find the probabilities of M throwing k heads and Chuck throwing k heads and then multiply them together because they are independent. Then sum those results.
$P(M≠C) = 1 - P(M=C)$
Similar expansion of possibilities gives the numerator.
A: We need to find probability of Mandy getting more H than Chuck given they both get different number of H. So our sample space reduces to possibilities where they get different number of H.
Number of possibilities that get same number of heads - either they both get no H, they both get $1$ H or they both get $2$ H.
$ = 1 + 2 \cdot 3 + 3 = 10$
So number of possibilities that they both get different number of H $ = 32 - 10 = 22$
Please note that possibilities of Mandy getting more H than Chuck is a subset of them getting different number of H. Here are the possibilities for Mandy to get more H than Chuck -
Chuck gets no H, then it is $7$ ways (other than Mandy getting zero H)
Chuck gets one H, then it is $2 \cdot 4$ ways (Mandy can get two or three H)
Chuck gets two H, then it is just $ \ 1$ way (Mandy must get three H)
So total number of possibilities for Mandy to get more H than Chuck $ = 7 + 2 \cdot 4 + 1 = 16$.
Desired probability $ = \frac{16}{22} = \frac{8}{11}$.
