A) \[\frac{1}{16}\]
B) \[\frac{5}{256}\]
C) \[\frac{35}{128}\]
D) None of these
Correct Answer: C
Solution :
Both players can get equal number of heads 0, 1, 2, 3 and 4. \[\therefore \]\[{{P}_{1}}\] (both players get 0 head) \[=\frac{1}{16}\times \frac{1}{16}\] \[{{P}_{2}}\] (both players get 1 head) \[=\frac{4}{16}\times \frac{4}{16}\] \[{{P}_{3}}\] (both players get 2 heads) \[=\frac{6}{16}\times \frac{6}{16}\] \[{{P}_{4}}\] (both players get 3 heads) \[=\frac{4}{16}\times \frac{4}{16}\] \[{{P}_{5}}\] (both players get 4 heads) \[=\frac{1}{16}\times \frac{1}{16}\] \[\therefore \]Required probability \[P={{P}_{1}}+{{P}_{2}}+{{P}_{3}}+{{P}_{4}}+{{P}_{5}}\] \[=\frac{1}{16\times 16}+\frac{16}{16\times 16}+\frac{36}{16\times 16}\] \[+\frac{16}{16\times 16}+\frac{1}{16\times 16}\] \[=\frac{1+16+36+16+1}{16\times 16}=\frac{70}{16\times 16}=\frac{35}{128}\]
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