\(Let\ K(n)\ be\ the\ probability\ that\ Keiko\ gets\ n\ heads,\ and\ let\ E(n)\ be\ the\ probability\ that\ Ephriam\ gets\\ \\ n\ heads. \\ \\ K(0) = 1/2 \\ \\K(1) = 1/2 \\ \\K(2) = 0\ (Keiko\ only\ has\ one\ penny!) \\ \\E(0) = \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4} \\ \\E(1) = \frac{1}{2}\cdot\frac{1}{2} + \frac{1}{2}\cdot\frac{1}{2} \\ \ \ \ \ \ \ \ \ = 2\cdot\frac{1}{4} \ \ \ \ \ \ \ \ = \frac{1}{2} \\ (because\ Ephraim\ can\ get\ HT\ or\ TH) \\ \\E(2) = \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4} \\ \\The\ probability\ that\ Keiko\ gets\ 0\ heads \\ \\ and\ Ephriam\ gets\ 0\ heads\ is\ K(0)\cdot E(0). \\ \\ Similarly\ for\ 1\ head\ and\ 2\ heads.\ Thus,\ we\ have: \\ \\P = K(0)\cdot E(0) + K(1)\cdot E(1) + K(2)\cdot E(2) \\ \\P = \frac{1}{2}\cdot\frac{1}{4} + \frac{1}{2}\cdot\frac{1}{2} + 0 \\ \\P = \frac{3}{8} \\ \\Thus\ the\ answer\ is\ 3/8.\)
\(-tommarvoloriddle\)
EDIT:
SUMMARY
- we used a equation K(0)*E(0)+K(1)+E(1)+K(2)*E(2)
- we found the values by using probability and logic.
- we plugged the numbers in
- we got the answer 3/8.
Apologises:
-quality of the answer
-all the work is in latex
-fact that it may not be super clear
If you have a question, just let me know.
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