Question

1. A coin is tossed 8 times.

PART I:

a) Find the chance of getting 8 heads in a row

b) Given that the first 7 tosses were heads, find the chance of getting 8 heads in a row

c) Given that the first 6 tosses were heads, find the chance of getting 8 heads in a row

PART II

a) Find the chance to get exactly 5 heads in 8 tosses (binomial formula).

b) Find the chance of getting at least one head in 8 tosses. Explain

Answer 1

1. When a Coin is toss

P( Head) = P(H) = 0.5

P( Tail) = P(T) = 0.5

PART I

a) P( 8 heads in a row) = (0.5)8 = 0.00391

b) P( 8 heads in a row given that the first 7 tosses were heads)

= P( 8th toss is a Head | first 7 tosses were heads)

= P( 8th toss is a Head n first 7 tosses were heads) / P( First 7 tosses were heads)

= (0.5)8 / (0.5)7

= 0.5

c)

P( 8 heads in a row given that the first 6 tosses were heads)

= P( 7th and 8th toss is a Head | first 6 tosses were heads)

= P( 7th and 8th toss is a Head n first 6 tosses were heads) / P( First 6 tosses were heads)

= (0.5)8 / (0.5)6

= 0.52

PART II

Let X be the number of heads in 8 tosses

X~ Binomial ( 8, 0.5)

a) P( X=5) = 8C5 (0.5)3 (1-0.5)5

=0.2187

b) P( X>=1) = 1- P(X<1)

= 1- P(X=0)

= 1-  8C0 (0.5)0 (1-0.5)8

= 1- 0.0039

= 0.9961