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  • 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 toses

7.    A fair coin is tossed four times, and a person win Re \small 1 for each head and lose Rs \small 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 (1)

best_answer

Here the sample space is,

S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, THHT, HTTH, THTH, TTHH, TTTH, TTHT, THTT, HTTT, TTTT}

According to question,

1.)  4 heads = 1 + 1 + 1 + 1 = Rs. 4 

2.)  3 heads and 1 tail = 1 + 1 + 1 - 1.50 = Rs. 1.50 

3.)  2 heads and 2 tails = 1 + 1 - 1.50 - 1.50 = - Rs. 1 : he will lose Re. 1

4.)  1 head and 3 tails = 1 – 1.50 – 1.50 – 1.50 = - Rs. 3.50 : he will lose Rs. 3.50

5.)  4 tails = – 1.50 – 1.50 – 1.50 – 1.50 = - Rs. 6 = he will lose Rs. 6

Now, sample space of amounts corresponding to S:

S' = {4, 1.50, 1.50, 1.50, 1.50, - 1, - 1, - 1, - 1, - 1, - 1, - 3.50, - 3.50, - 3.50, - 3.50, - 6}

\therefore n(S') = 12

\therefore Required Probabilities are: 

P(Winning\ Rs.\ 4) = \frac{n(Winning\ Rs.\ 4)}{n(S')}  = \frac{1}{16}

P(Winning\ Rs.\ 1.50) = \frac{n(Winning\ Rs.\ 1.50)}{n(S')}  = \frac{4}{16} = \frac{1}{4}

P(Losing\ Re.\ 1) = \frac{n(Losing\ Re.\ 1)}{n(S')}  = \frac{6}{16} = \frac{3}{8}

P(Losing\ Rs.\ 3.50) = \frac{n(Losing\ Rs.\ 3.50)}{n(S')}  = \frac{4}{16} = \frac{1}{4}

P(Losing\ Rs.\ 6) = \frac{n(Losing\ Rs.\ 6)}{n(S')}  = \frac{1}{16}

Posted by

HARSH KANKARIA

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