Asked: What is the probability of getting four tails and two heads when an unfair coin is tossed 6 times?
Let us assume that probability of head =p
Probability of tail =1-p
The probability of getting four tails and two heads when an unfair coin is tossed 6 times = 6C2×p^2(1-p)^4=15p^2(1-p)^4
(1) When the unfair coin is tossed thrice, the probability of getting two tails and one heads is 2/9.
The probability of getting two tails and one heads = 3C1×p×(1-p)^2= 3p(1-p)^2=2/9.
p=2/3; 1-p=1/3
The probability of getting four tails and two heads when an unfair coin is tossed 6 times = 6C2×p^2(1-p)^4=15p^2(1-p)^4=20/243
SUFFICIENT
(2) When the unfair coin is tossed twice, the probability of getting one tails and one heads is 4/9.
The probability of getting one tails and one heads = 2C1×p×(1-p)=2p(1-p)=4/9.
p=2/3; 1-p=1/3
The probability of getting four tails and two heads when an unfair coin is tossed 6 times = 6C2×p^2(1-p)^4=15p^2(1-p)^4=20/243
SUFFICIENT
IMO D
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Kinshook Chaturvedi
Email: kinshook.chaturvedi@gmail.com