# Probability Distrubtion - ANSWER CHECK!

• Jan 5th 2009, 05:27 PM
Last_Singularity
Quote:

Originally Posted by cnmath16
48. A coin is tossed 8 times. Calculate the probability of tossing

a) 6 heads and 2 tails

b) 7 heads and 1 tail

The coin question is binomially distributed with $p = 0.5$ and $n = 8$. The problem of getting $k$ successes out of $n$ trials is calculated by
$\frac{n!}{(n-k)!k!} p^k (1-p)^{n-k}$

For more properties of the distribution, see: Binomial distribution - Wikipedia, the free encyclopedia
• Jan 8th 2009, 05:37 PM
cnmath16
48. A coin is tossed 8 times. Calculate the probability of tossing..

a) 6 heads and 2 tails

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
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

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 PM
mr fantastic
Quote:

Originally Posted by cnmath16
48. A coin is tossed 8 times. Calculate the probability of tossing..

a) 6 heads and 2 tails

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
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