Question

A fair coin is tossed four times, and a person wins Re 1 for each head and losses Rs.1.50 for each trail 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

Hint: Write all the cases possible one by one and then find profit on falling head and also the loss on falling tail. Subtract them profit – loss is nothing but the amount of money they will have in each case. Find the number of possibilities of each case. Dividing this with total possibilities gives us the probability of getting each amount.
Amount = (number of heads) \[\times \] 1 – (number of tails) \[\times \] 1.5

Complete step-by-step answer:

When a fair coin is tossed 4 times we have possibilities each time: a head or a tail.
So, we have a total of 5 cases possible here.
Case 1: 4 heads 0 tails in the 4 trails they throw.
Case 2: 3 heads 1 tail in the 4 trails they throw.
Case 3: 2 heads 2 tails in the 4 trails they throw.
Case 4: 1 heads 3 tails in the 4 trails they throw.
Case 5: 0 heads 4 tails in the 4 trails they throw.
Now we need the amount they have after each case is completed. For that we have information each head is profit of Re 1 each tail is loss of 1.5 Re.
By thus we can say: -
Total amount = (Head count) \[\times \] 1 – (Tail count) \[\times \] 1.5.
Calculating amount for each case,
Case 1: (4). (1) – 0
By simplifying we can say that the amount in the case of 4 heads is 4 Re.
Case 2: (3). (1) – (1) (1.5)
By simplifying we can say that the amount in the case of 3 heads is 1.5 Re.
Case 3: (2). (1) – (2) (1.5)
By simplifying we can say that the amount in the case of 2 heads is -1 Re.
Case 4: (1) (1) – (3) (1.5)
By simplifying we can say that the amount in the case of 1 head is -3.5 Re.
Case 5: (0) (1) – (4) (1.5)
By simplifying we can say that the amount in the case of 0 heads is -6 Re.
Hence the different amounts of money are 4, 1.5, -1, -3.5, -6.
Each time a coin is thrown we have 2 possibilities.
So, total possibilities = 2 \[\times \] 2 \[\times \] 2 \[\times \] 2 = \[{{2}^{4}}\].
Total possibilities = 16.
Calculating probability of each case,
Case 1: 4 heads 0 tails
HHHH \[\Rightarrow \]1 possibility, probability \[=\dfrac{1}{16}\].
Case 2: 3 heads 1 tail \[=\dfrac{4!}{3!}\]
HHHH \[\Rightarrow \] 4 possibilities, probability \[=\dfrac{1}{4}\].
Case 3: 2 heads 2 tails
\[\Rightarrow \dfrac{4!}{2!2!}\] = 6 possibilities, probability \[=\dfrac{3}{8}\].
Case 4: 1 heads 3 tails
\[\Rightarrow \dfrac{4!}{3!}\] = 4 possibilities, probability \[=\dfrac{1}{4}\].
Case 5: 0 heads 4 tails
\[\Rightarrow \dfrac{4!}{4!}\] = 1 possibility, probability \[=\dfrac{1}{16}\].
Hence these are probabilities for each case.

Note: It is important to consider all five cases by thinking properly , if we miss out on any of the cases then the complete solution would be wrong. Also keep in mind that whenever n number of coins are tossed simultaneously the total possible outcomes will be ${n}^2$.