Question

A fair coin is tossed four times, and a person wins Re 1 for each head and loses Rs. $1.50$ for each tail that turns up. From the sample space calculate how many different amounts of money you can have after four tosses and the probability of having each of these amounts.

Answer 1

Given a coin is tossed four times

∴∴ Possible outcomes =2*2*2*2=1624=16

∴∴ Sample space can be written as

Sample space =HHHH, Amount =1+1+1+1=4

Sample space =HHHT, Amount =1+1+1-1.50=3-1.50=1.50

Sample space =HHTH, Amount =1+1-1.50+1=1.50

Sample space =HHTT, Amount =1+1-1.50-1.50=-1.00

Sample space =HTHH, Amount =1-1.50+1+1=1.50

Sample space =HTHT, Amount =1-1.50+1-1.50=-1.00

Sample space =HTTH, Amount =1-1.50-1.50+1=-1.00

Sample space =HTTT, Amount =1-1.50-1.50-1.50=-3.50

Sample space =THHH, Amount =-1.50+1+1+1=1.50

Sample space =THHT, Amount =-1.50+1+1-1.50=-1.00

Sample space =THTH, Amount =-1.50+1-1.50+1=-1.00

Sample space =THTT, Amount =-1.50+1-1.50-1.50=-3.50

Sample space =TTHH, Amount =-1.50-1.50+1+1=-1.00

Sample space =TTHT, Amount =-1.50-1.50+1-1.50=-3.50

Sample space =TTTH, Amount =-1.50-1.50-1.50+1=-3.50

Sample space =TTTT, Amount =-1.50-1.50-1.50-1.50=-6

Therefore from the samples we get 5 type of different amounts.

4,1.50,−1.00,−3.50,−6.004,1.50,−1.00,−3.50,−6.00

4 has occurred 1 time

1.50 has occurred 4 times

-1.00 has occurred 6 times

-3.50 has occurred 4 times

-6.00 has occurred 1 time

⇒⇒ P(winning of Rs.4.00)=Number of favorable outcomesTotal number of outcomesNumber of favorable outcomesTotal number of outcomes

⇒116⇒116

⇒⇒P(winning of Rs1.50)=4/16=1/4
416=14

⇒⇒P(winning of Rs-1.00)=6/16=3/8616=38

⇒⇒P(winning of Rs-3.50)=4/16=1/4
416=14

⇒⇒P(winning of Rs-6.00)=1/16