Question

A fair coin is tossed, and a fair die is rolled. Let H be the event that the coin lands on heads, and let S be the event that the die lands on six. Find P(H or S).

Answer 1

Solution:

Given: A fair coin is tossed, and a fair die is rolled.

Thus sample space of tossing a fair coin and a fair die is:

= { H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6 }

N = total number of outcomes in sample space = 12

H be the event that the coin lands on heads

H : { H1, H2, H3, H4, H5, H6 }

n(H) = number of outcomes H = 6

S be the event that the die lands on six

S : { H6,    T6 }

n(S) = Number of outcomes of die lands on six = 2

and

H and S : { H6 }

n( H and S) = Number of outcomes on which coin lands on head and die lands on six = 1

We have to find:

P( H or S) = ...........?

Using addition rule of probability:

P( H or S) = P(H) + P(S) - P( H and S)

P( H ) = n(H) / N = 6 / 12

P( S ) = n(S) / N = 2 / 12

and

P( H andS) = n(H and S) /N = 1/12

thus

P( H or S) = P(H) + P(S) - P( H and S)

P( H or S) = 6/12 + 2/12 - 1/12

P( H or S) = ( 6 + 2 - 1 ) / 12

P( H or S) = 7 / 12

P( H or S) = 0.5833