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प्रश्न
A and B throw a pair of dice. If A throws 9, find B's chance of throwing a higher number.
उत्तर
A and B throw a pair of dices.
Then all the possible outcomes = 62 = 36
i.e. total number of possible outcome, n(S) = 36
Consider E = event where A throws 9 and B throws more than 9, i.e. 10, 11 and 12
E = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}
i.e. n(E) = 6
Hence, required probability = \[\frac{6}{36} = \frac{1}{6}\]
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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` |
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.
Three coins are tossed once. Find the probability of getting
- 3 heads
- 2 heads
- at least 2 heads
- at most 2 heads
- no head
- 3 tails
- exactly two tails
- no tail
- atmost two tails.
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Check whether the following probabilities P(A) and P(B) are consistently defined
P(A) = 0.5, P(B) = 0.4, P(A ∪ B) = 0.8
Fill in the blank in following table:
| P(A) | P(B) | P(A ∩ B) | P(A ∪ B) |
| `1/3` | `1/5` | `1/15` | .... |
Fill in the blank in following table:
| P(A) | P(B) | P(A ∩ B) | P(A ∪ B) |
| 0.35 | ... | 0.25 | 0.6 |
Fill in the blank in following table:
| P(A) | P(B) | P(A ∩ B) | P(A ∪ B) |
| 0.5 | 0.35 | .... | 0.7 |
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| S. No. | Name | Sex | Age in years |
| 1. | Harish | M | 30 |
| 2. | Rohan | M | 33 |
| 3. | Sheetal | F | 46 |
| 4. | Alis | F | 28 |
| 5. | Salim | M | 41 |
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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 |
| (ii) |
\[\frac{1}{7}\]
|
\[\frac{1}{7}\]
|
\[\frac{1}{7}\]
|
\[\frac{1}{7}\]
|
\[\frac{1}{7}\]
|
\[\frac{1}{7}\]
|
\[\frac{1}{7}\]
|
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 |
| (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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