Thread: Probability 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

Originally Posted by cnmath16

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

All correct.