Question 47 of 55advanced🔧 ApplyLong Answer5 marks

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

Correct Answer

Step 1: Sample Space

When a fair coin is tossed 4 times, each toss has 2 equally likely outcomes (Head or Tail). The total number of outcomes in the sample space:

=24=16= 2^4 = 16

The outcomes can be classified based on number of heads:

  • 4 heads, 0 tails: HHHH (1 outcome)
  • 3 heads, 1 tail: HHHT, HHTH, HTHH, THHH (4 outcomes)
  • 2 heads, 2 tails: HHTT, HTHT, HTTH, THHT, THTH, TTHH (6 outcomes)
  • 1 head, 3 tails: HTTT, THTT, TTHT, TTTH (4 outcomes)
  • 0 heads, 4 tails: TTTT (1 outcome)

Step 2: Calculating Different Amounts

Given: Win Re 1 for each head, Lose Rs 1.50 for each tail.

  • 4 heads, 0 tails:

Amount=4×1−0×1.50=Rs 4\text{Amount} = 4 \times 1 - 0 \times 1.50 = \textbf{Rs 4}

  • 3 heads, 1 tail:

Amount=3×1−1×1.50=Rs 1.50\text{Amount} = 3 \times 1 - 1 \times 1.50 = \textbf{Rs 1.50}

  • 2 heads, 2 tails:

Amount=2×1−2×1.50=Rs (-1)\text{Amount} = 2 \times 1 - 2 \times 1.50 = \textbf{Rs (-1)}

  • 1 head, 3 tails:

Amount=1×1−3×1.50=Rs (-3.50)\text{Amount} = 1 \times 1 - 3 \times 1.50 = \textbf{Rs (-3.50)}

  • 0 heads, 4 tails:

Amount=0×1−4×1.50=Rs (-6)\text{Amount} = 0 \times 1 - 4 \times 1.50 = \textbf{Rs (-6)}

Number of different amounts = 5

Step 3: Probability of Each Amount

Using:

P(E)=Number of favourable outcomesTotal outcomesP(E) = \frac{\text{Number of favourable outcomes}}{\text{Total outcomes}}

  • P(Amount = Rs 4):

=116= \frac{1}{16}

  • P(Amount = Rs 1.50):

=416=14= \frac{4}{16} = \frac{1}{4}

  • P(Amount = Rs -1):

=616=38= \frac{6}{16} = \frac{3}{8}

  • P(Amount = Rs -3.50):

=416=14= \frac{4}{16} = \frac{1}{4}

  • P(Amount = Rs -6):

=116= \frac{1}{16}

Verification:

Sum of probabilities=116+416+616+416+116=1616=1\text{Sum of probabilities} = \frac{1}{16} + \frac{4}{16} + \frac{6}{16} + \frac{4}{16} + \frac{1}{16} = \frac{16}{16} = 1

Exercise: EXERCISE 14.2 | Q: 7 | (Chapter: Page 17)
For More Understanding

Explanation

This question tests understanding of sample space construction and probability calculation. The key insight is that outcomes are classified by the number of heads (or tails), not the specific sequence.

Since each outcome in the sample space is equally likely (probability 116\frac{1}{16}), we count how many outcomes correspond to each amount. The binomial pattern (1,4,6,4,1)(1, 4, 6, 4, 1) emerges naturally from counting combinations of heads and tails.

Solution Steps

  1. Step 1: Determine sample space size = 24=162^4 = 16 equally likely outcomes

  2. Step 2: Classify outcomes by number of heads (4H:1, 3H:4, 2H:6, 1H:4, 0H:1)

  3. Step 3: Calculate amount for each case using formula: Amount=(heads×1)−(tails×1.50)\text{Amount} = (\text{heads} \times 1) - (\text{tails} \times 1.50)

  4. Step 4: Find probability of each amount by dividing favourable outcomes by total outcomes

  5. Step 5: Verify that sum of all probabilities equals 1

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