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प्रश्न
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.
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उत्तर
There are five ways in which a head can be obtained in a coin toss. These are as follows.
Total possible outcomes = {HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTT, TTHH, TTHT, TTTH, TTTT}
(i) No head is obtained or all four tails are obtained.
Loss on getting all four tails = 4 × 1.50
= Rs. 6
Ways of getting four tails (TTTT) = 1
Total possible outcomes = 16
∴ Probability of getting four tails = `1/16`
(ii) When one head and 3 tails are obtained.
Loss = 3 × 1.50 – 1 × 1
= 4.50 – 1.00
= Rs. 3.50
A head and 3 tails can come up as follows:
{TTTH, TTHT, THTT, HTTT}
∴ A head and 3 tails can come up in 4 ways.
Total possible outcomes = 16
Probability of getting a head = `6/16`
= `1/4`
(iii) When 2 heads and 2 tails appear
Loss = 2 × 1.5 – 1 × 2
= 3 – 2
= Rs. 1
2 heads and 2 tails can come up as follows.
{HHTT, HTHT, HTTH, THHT, THTH, TTHH}
There are six ways in which 2 heads and 2 tails can be obtained.
Total possible outcomes = 16
Probability of getting 2 heads = 2
(iv) When 3 heads and 1 tail appear, then
Profit = 3 × 1 – 1 × 1.5
= 3 – 1.50
= Rs. 1.50
Ways of getting 3 heads = {HHHT, HHHH, HTHH, THHH}
There are four ways in which 3 heads and 1 tail can be obtained.
Total possible outcomes = 16
Probability of getting 3 heads = `4/16`
= `1/4`
(v) All four heads can be obtained in one way, then
Profit = 4 × 1
= Rs. 4
Total possible outcomes = 16
Probability of getting four heads = `4/16`
= `1/4`
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संबंधित प्रश्न
Which of the following can not be valid assignment of probabilities for outcomes of sample space S = {ω1, ω2,ω3,ω4,ω5,ω6,ω7}
| Assignment | ω1 | ω2 | ω3 | ω4 | ω5 | ω6 | ω7 |
| (a) | 0.1 | 0.01 | 0.05 | 0.03 | 0.01 | 0.2 | 0.6 |
| (b) | `1/7` | `1/7` | `1/7` | `1/7` | `1/7` | `1/7` | `1/7` |
| (c) | 0.1 | 0.2 | 0.3 | 0.4 | 0.5 | 0.6 | 0.7 |
| (d) | –0.1 | 0.2 | 0.3 | 0.4 | -0.2 | 0.1 | 0.3 |
| (e) | `1/14` | `2/14` | `3/14` | `4/14` | `5/14` | `6/14` | `15/14` |
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| `1/3` | `1/5` | `1/15` | .... |
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| 0.5 | 0.35 | .... | 0.7 |
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| Elementary events: | w1 | w2 | w3 | w4 | w5 | w6 | w7 |
| (iii) | 0.7 | 0.06 | 0.05 | 0.04 | 0.03 | 0.2 | 0.1 |
Which of the cannot be valid assignment of probability for elementary events or outcomes of sample space S = {w1, w2, w3, w4, w5, w6, w7}:
| Elementary events: | w1 | w2 | w3 | w4 | w5 | w6 | w7 |
| (iv) |
\[\frac{1}{14}\]
|
\[\frac{2}{14}\]
|
\[\frac{3}{14}\]
|
\[\frac{4}{14}\]
|
\[\frac{5}{14}\]
|
\[\frac{6}{14}\]
|
\[\frac{15}{14}\]
|
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