The coin question is binomially distributed with and . The problem of getting successes out of trials is calculated by

For more properties of the distribution, see: Binomial distribution - Wikipedia, the free encyclopedia

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- Jan 5th 2009, 05:27 PMLast_Singularity
The coin question is binomially distributed with and . The problem of getting successes out of trials is calculated by

For more properties of the distribution, see: Binomial distribution - Wikipedia, the free encyclopedia - Jan 8th 2009, 05:37 PMcnmath16Probability Distrubtion - ANSWER CHECK!
**48. A coin is tossed 8 times. Calculate the probability of tossing..**

**a) 6 heads and 2 tails**

My answer:

Let p=P(heads) = 1/2

Let q= P (tails) =1/2

From the expansion of (p+q)^8, use the term containing p^6.

Thus, P(6 heads) = c (8,2)(1/2)^6 (1/2)^2

= 8! ÷ 6! 2! (1/2)^8

= (56/2)(1/256)

= 28/256

= 14/128

= 7/64 is the probability of 6 heads and 2 tails occuring

**b) 7 heads and 1 tail**

My answer:

From the expansion of (p+q)^8, use the term containing p^7

Thus, P(7 heads) = c (8,1) (1/2)^7(1/2)

= 8! ÷ 7! 1! (1/2)^8

= 8/256

= 1/32 is the probability of 7 heads and 1 tail

**c) at least 6 heads**

My answer:

P (at least 6 heads) = P(6 heads) + P (7 heads) + P (8 heads)

P (6 heads) = 28/256

P (7 heads) = 8/256

P(8 heads) = 1/256

Thus, the total probability of 6 heads is equal to 28/256 + 8/256 + 1/256 which is equal to P= 37/256 - Jan 8th 2009, 06:41 PMmr fantastic