A fair coin is tossed four times, and people 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 amount

Options

A. `1/9`
B. `1/16`
C. `1/8`
D. `1/24`

Answer

C

Explanation

When a coin is tossed 4 times
Sample space
`S={{:(HHHH),(HHHT","HHTH","HTHH","THHH),(HHTH","HTHT","H"TT"H","THHT","THTH",""TT"HH),(H"TT"","TH"TT"",""TT"HT",""TT"TH),("TTTT"):}}`
`impliesn(S)=16`
Let event `A=` getting four heads
`=` winning Rs. 4.00
Event `B=` getting three heads and one tails
`=Rs. (3xx1.00-1xx1.50)`
`=` winning Rs. 1.50
Event `C=` getting two heads and two tails
`=Rs. (2xx1.00-2xx1.50)`
`=` losing Rs. 1.00
Event `D=` getting one head and three tails
`=Rs. (1xx1.00-3xx1.50)`
`=` losing Rs. 3.50
Event `E=` getting four tails
`=Rs. 4xx1.50=` losing Rs. 6.00
Now `n(A)=1,n(B)=4,n(C)=6`
`(D)=4` and `n(E)=1`
`:. P(A)=(n(A))/(n(S))=1/16`
`implies` Probability of winning Rs. 4.00 `=1/16`
Applying `P(B)=(n(B))/(n(S))=4/16=1/4`
`implies` Probability of winning Rs. `1.50=1/4`
`P(C)=(n(C))/(n(S))=6/16=3/8`
`implies` Probability of lossing Rs. 1.00=3/8`
`P(D)=(n(D))/(n(S))=4/16=1/4`
Probability of lossing Rs. `3.50=1/4`
`P(E)=(H(E))/(H(S))=1/16`
`implies` Probability of losing Rs. `6.00=1/16`