probability

[joytlq]

Question:
garfield and oldie each toss 3 fair coins. prove that
the probability that oldie gets more head than
garfield is 11/32

Heres my working for the qn after considering all the possible cases
P( oldie with 2Heads, 1 tail and garfield with 1head, 2 tails)
p(oldie with 2Heads, 1 tail and garfield with 0 head,3 tails)
p(oldie with 3Heads,0 tail and garfield with 1 head,2 tails)
p(oldie with 3Heads,0 tail and garfield with 2 heads,1 tail)
p(oldie with 3Heads,0 tail and garfield with 0 heads,3tails)
p(oldie with 1Heads,2 tails and garfield with 0 heads,3tails)
These are all the possible cases.
each case correspond to
(1/2)(1/2)(1/2)(1/20(1/2)(1/2)
hence, total probability is ((1/2)to the power of six) X 6 = 8/32
WhaT went wrong?
Thank you!

 

[galactus]

odie gets more head

A very lucky Odie in more ways than just tossing coins :lol:

Heres my working for the qn after considering all the possible cases
P( oldie with 2Heads, 1 tail and garfield with 1head, 2 tails)
p(oldie with 2Heads, 1 tail and garfield with 0 head,3 tails)
p(oldie with 3Heads,0 tail and garfield with 1 head,2 tails)
p(oldie with 3Heads,0 tail and garfield with 2 heads,1 tail)
p(oldie with 3Heads,0 tail and garfield with 0 heads,3tails)
p(oldie with 1Heads,2 tails and garfield with 0 heads,3tails)
These are all the possible cases.
each case correspond to
(1/2)(1/2)(1/2)(1/20(1/2)(1/2)
hence, total probability is ((1/2)to the power of six) X 6 = 8/32
WhaT went wrong?
Thank you!

You can use the binomial probability.

Three cases:

Odie gets one head, then Garfield gets none:

\(\displaystyle \binom{3}{1}(\frac{1}{2})^{1}(\frac{1}{2})^{2}\cdot \binom{3}{0}(\frac{1}{2})^{0}(\frac{1}{2})^{3}=\frac{3}{64}\)

Odie gets 2 heads, then Garfield gets 0 or 1 head:

\(\displaystyle \binom{3}{2}(\frac{1}{2})^{2}(\frac{1}{2})^{1}\cdot \sum_{k=0}^{1}\binom{3}{k}(\frac{1}{2})^{k}(\frac{1}{2})^{3-k}=\frac{3}{16}\)

Odie gets three heads, then Garfield gets 0,1, or 2 heads:

\(\displaystyle \binom{3}{3}(\frac{1}{2})^{3}(\frac{1}{2})^{0}\cdot \sum_{k=0}^{2}\binom{3}{k}(\frac{1}{2})^{k}(\frac{1}{2})^{3-k}=\frac{7}{64}\)

\(\displaystyle \frac{7}{64}+\frac{3}{16}+\frac{3}{64}=\frac{11}{32}\)

 

[soroban]

Hello, joytlq!

Garfield and Odie each toss 3 fair coins.
Prove that the probability that Odie gets more Heads than Garfield is 11/32

Odie tosses 3 coins.
. . There are \(\displaystyle 2^3 \,=\,8\) possible outcomes: .HHH, HHT, HTH, HTT, THH, THT, TTH, TTT

Garfield tosses 3 coins.
. . He has the same 8 possible outcomes.

Together, there are: \(\displaystyle 8 \times 8 \,=\,64\) possible outcomes.


Consider the cases in which they have an equal number of Heads.

. . \(\displaystyle \begin{array}{|cc|cc||c|} \text{Garfield } & \text{Prob} & \text{Odie} & \text{Prob} & P(\text{both}) \\ \hline 3H & \frac{1}{8} & 3H & \frac{1}{8} & \frac{1}{64} \\ \\[-3mm] 2H,1T & \frac{3}{8} & 2H,1T & \frac{3}{8} & \frac{9}{64} \\ \\[-3mm] 1H,2T & \frac{3}{8} & 1H,2T & \frac{3}{8} & \frac{9}{64} \\ \\[-3mm] 3T & \frac{1}{8} & 3T & \frac{1}{8} & \frac{1}{64} \\ \hline \end{array}\)

\(\displaystyle \text{Then: }\(\text{equal H's}) \;=\;\frac{1}{64} + \frac{9}{64} + \frac{9}{64} + \frac{1}{64} \:=\:\frac{20}{64}\;=\;\frac{5}{16}\)


The rest of the time they have unequal numbers of Heads.

. . \(\displaystyle P(\text{unequal H's}) \;=\;1 - \frac{5}{16} \:=\:\frac{11}{16}\)


And in half of those cases, Odie will have more Heads than Garfield.

. . \(\displaystyle P(\text{Odie} > \text{Garfield}) \;=\;\frac{1}{2}\cdot\frac{11}{16} \;=\;\frac{11}{32}\)