Question

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.

Answer 1

In the given experiment, a fair coin is tossed four times. A person wins 1 rupee each time a head (H) turns up and loses 1.50 rupee each time a tail (T) turns up.

The various possible outcomes are as follows,

Let, event A: All four heads. The amount with man is given by,

Re 1+Re 1+Re 1+Re 1=Rs 4( gain )

Let, event B: All four tails. The amount with man is given by,

−Rs 1.5−Rs 1.5−Rs 1.5−Rs 1.5=−Rs 6( loss )

Let, event C: Two head, two tails. The amount with man is given by, −Rs 1.5−Rs 1.5+Re 1+Re 1=−Re 1( loss )

Let, event D: Three heads, one tail. The amount with man is given by, Re 1+Re 1+Re 1−R 1.5=Rs 1.5( gain )

Let, event E: Three tails, one head. The amount with man is given by, −Rs 1.5−Rs 1.5−Rs 1.5+Re 1=−Rs 3.5( loss )

The sample space, S, consists of sixteen elements: S={ ( H,H,H,H ),( T,T,T,T ),( H,H,T,T ),( H,T,H,T ), ( H,T,T,H ),( T,H,T,H ),( T,T,H,H ),( T,H,H,T ), ( H,H,H,T ),( H,H,T,H ),( H,T,H,H ),( T,H,H,H ), ( T,T,T,H ),( T,T,H,T ),( T,H,T,T ),( H,T,T,T ) }

n( S )=16

The probability for event A “gaining rupees 4”is,

P( A )= number of outcomes favourable to A Total number of outcomes = n( A ) n( S ) = 1 16

Thus, the probability for event A “gaining rupees 4” is 1 16 .

The probability for event B “losing rupees 6” is,

P( B )= number of outcomes favourable to B Total number of outcomes = n( B ) n( S ) = 1 16

Thus, the probability for event B “losing rupees 6” is 1 16 .

The probability for event C “losing rupee 1” is,

P( C )= number of outcomes favourable to C Total number of outcomes = n( C ) n( S ) = 6 16 = 3 8

Thus, the probability for event C “losing rupee 1” is 3 8 .

The probability for event D “gaining rupees 1.5“is,

P( D )= number of outcomes favourable to D Total number of outcomes = n( D ) n( S ) = 4 16 = 1 4

Thus, the probability for event D “gaining rupees 1.5” is 1 16 .

The probability for event E “losing rupees 3.5” is,

P( E )= number of outcomes favourable to E Total number of outcomes = n( E ) n( S ) = 4 16 = 1 4

Thus, the probability for event E “losing rupees 3.5” is 1 16 .