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A fair coin with sides marked heads and tails is to be [#permalink]

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02 Mar 2005, 11:16

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

A fair coin with sides marked heads and tails is to be tossed eight times. What is the probability that the coin will land tails side up more than five times?Please explain this for me.

Probability of getting tails eight times = 1/256
Probability of getting tails seven times = 1/128
Probability of getting tails six times = 1/64
So Probability of getting tails more than five times = 1/64 + 1/128 + 1/256 = 7/256

8 times => 8c8 * 256 = 1/256 7 times => 8c7 * 256 = 8/256 6 times => 8c6 * 256 = 28 / 256

=> 37/256

I totally agree with christoph
You have (1/2)^8 ways of getting a different result (H or T). Then with each of the outcomes(>5... which is 6,7 and 8), you have 8C6, 8C7 and 8C8 ways of sorting the H possibilities. _________________

Don't be so harsh on yourself. You just wasn't as used to the formula as I am. From now on you'll be always right on this type of questions, won't you?

Fair Coin -------> probablity of heads or tails = 1/2

You need to pick more than 5 instances out of 8 where it is heads/tails(it doesnt matter as it is a fair coin)

So it should be 6,7 or 8 times the desired outcome may arrive

Take 6 times out of 8 then the chaces of heads/tails
= 8C6 (1/2)^6*(1/2)^2

that is nCr*P(E)^a*P(E!)^b

Where n = no of total outcomes
r = no of desired outcomes
P(E) is probablity of desired outcomes
a -------> no of desired outcomes
P(E!) is probablity of UNdesired outcomes
b -------> no of UNdesired outcomes

n=a+b

Since we have r=6,7,8 add them all after putting in the above formula

A fair coin with sides marked heads and tails is to be tossed eight times. What is the probability that the coin will land tails side up more than five times?Please explain this for me.

Prob of 8 tails= C(8,8)*0.5^8
Prob of 7 tails= C(8,7)*0.5^7*0.5
Prob of 6 tails= C(8,6)*0.5^6*0.5^2