We have 5 coins and each can be tail or head == > 5^2 = 32 possible states
getting at most 3 tails means 1, 2 or 3 tails, in other words 4 or 5 heads.
how many states with 5 heads ==>1
how many states with 4 heads ==>5 |=> 5+1 =6 ==> 32-6=26 states with 1, 2 or 3 tails,
Probability = 26/32= 13/16
Probability Question
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sanjoy18
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probability of getting at most 3 tails= 1- probability of getting at least 4 tails
probability of getting at least 4 tails
=probability of getting 4 tails + probability of getting 5 tails
= 5/2^5 +1/2^5
= 6/32
hence required probability= 1-6/32= 26/32= 13/16
[NOTE: probability of getting 4 tails..their would be 5 cases TTTTH,TTTHT,TTHTT,THTTT,HTTTT]
probability of getting at least 4 tails
=probability of getting 4 tails + probability of getting 5 tails
= 5/2^5 +1/2^5
= 6/32
hence required probability= 1-6/32= 26/32= 13/16
[NOTE: probability of getting 4 tails..their would be 5 cases TTTTH,TTTHT,TTHTT,THTTT,HTTTT]
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sanjoy18
- Senior | Next Rank: 100 Posts
- Posts: 81
- Joined: Tue Jun 11, 2013 10:24 pm
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This problem can be solved by using binomial distribution formula
p(x)=(ncx)*p^x q^(1-x)
where p =probability getting tail=1/2
q=1-p=1/2
according to the problem, we have to find out
P(x<=3)=p(x=0)+p(x=1)+p(x=2)+p(x=3)-----------------(A)
=(1/32)*[5c0+5c1+5c2+5c3]
=26/32=13/16 (ANS)
(A) can be written as
p(x<=3)=1-[p(x=4)+p(x=5)]
=1-6/32
=13/16
p(x)=(ncx)*p^x q^(1-x)
where p =probability getting tail=1/2
q=1-p=1/2
according to the problem, we have to find out
P(x<=3)=p(x=0)+p(x=1)+p(x=2)+p(x=3)-----------------(A)
=(1/32)*[5c0+5c1+5c2+5c3]
=26/32=13/16 (ANS)
(A) can be written as
p(x<=3)=1-[p(x=4)+p(x=5)]
=1-6/32
=13/16
