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Question: A fair coin is tossed a fixed number of times. If the probability of getting 4 heads is equal to the...

A fair coin is tossed a fixed number of times. If the probability of getting 4 heads is equal to the probability of getting 7 heads, then the probability of getting 2 heads is

A

1/1024

B

3/4096

C

55/2048

D

None

Answer

55/2048

Explanation

Solution

If the coin is tossed n times, then

nC4(12)n=nC7(12)n{ } ^ { \mathrm { n } } \mathrm { C } _ { 4 } \left( \frac { 1 } { 2 } \right) ^ { \mathrm { n } } = { } ^ { \mathrm { n } } \mathrm { C } _ { 7 } \left( \frac { 1 } { 2 } \right) ^ { \mathrm { n } }

Ž nC4=nC7{ } ^ { \mathrm { n } } \mathrm { C } _ { 4 } = { } ^ { \mathrm { n } } \mathrm { C } _ { 7 }Ž n = 11

and then P(2 heads)

= nC2(12)2(12)n−2=nC2(12)n{ } ^ { \mathrm { n } } \mathrm { C } _ { 2 } \left( \frac { 1 } { 2 } \right) ^ { 2 } \left( \frac { 1 } { 2 } \right) ^ { \mathrm { n } - 2 } = { } ^ { \mathrm { n } } \mathrm { C } _ { 2 } \left( \frac { 1 } { 2 } \right) ^ { \mathrm { n } }

= 11C2(12)11=55(12)11=552048{ } ^ { 11 } \mathrm { C } _ { 2 } \left( \frac { 1 } { 2 } \right) ^ { 11 } = 55 \left( \frac { 1 } { 2 } \right) ^ { 11 } = \frac { 55 } { 2048 }