Question

if I need to know what is the probability of having 7, 8, 9, 10,11 heads in a row when a fair coin toss 70 times? Please explain the steps on how to calculate it. I would like to thank you in advance for helping me.

Answer 1

X follow binomial distribution with paramters n and p

P(X = k) = nCk p^k q^(n-k)

when n is large

it is not easy to calculate using above formula

we use normal distribution to approximate it

mean = np

sd =sqrt(npq)

P(X = k) = P(k-0.5 <X < k +0.5)

Z = (X - np)/sqrt(npq)

now

here n = 70 , p = 0.5

suppose k = 7

hence

P( X = 7) = P(6.5 <X< 7.5)

Z = (X - 70*0.5)/sqrt(70*0.5*0.5)

P(6.5 <X< 7.5)

= P ( −6.81<Z<−6.57 )=0

if we want

P(X = 7,8,9,10,11)

= P(6.5 < X< 11.5)

= P ( −6.81<Z<−5.62 )=0