Math, asked by bhaskarkalawat4766, 1 year ago

A fair coin is tossed four times, and a person win Re 1 for each head and lose 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.

Answers

Answered by amitnrw
9

Answer:

5 different amounts of money

Step-by-step explanation:

A fair coin is tossed four times

Total number of possibilties = 2⁴ = 16

T1   T2   T3  T 4  Net Won Net Loss

H H H H  4

H H H T 1.5

H H T H 1.5

H H T T             1

H T H H 1.5

H T H T              1

H T T H              1

H T T T           3.5

T H H H 1.5

T H H T          1

T H T H          1

T H T T        3.5

T T H H          1

T T H T         3.5

T T T H         3.5

T T T T         6

5 Different amount we Can have

0 - Head    Loss  = 6      Probability = 1/16

1 - Head    Loss = 3.5     Probability = 4/16  = 1/4

2 - Head   Loss = 1         Probability = 6/16   = 3/8

3 Head     Won = 1.5      Probability = 4/16  = 1/4

4 Head     Won =  4         Probability = 1/16

Answered by GangsterTeddy
1

Step-by-step explanation:

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

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