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प्रश्न
A box contains 100 bulbs, 20 of which are defective. 10 bulbs are selected for inspection. Find the probability thatat least one is defective
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उत्तर
Out of 100 bulbs, 10 can be chosen in 100C10 ways.
So, total number of elementary events = 100C10
Probability for at least one defective bulb = 1 – Probability (all 10 are non-defective)
= \[1 - \frac{^{80}{}{C}_{10}}{^{100}{}{C}_{10}}\]
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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 die is thrown, find the probability of following events:
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- A number greater than or equal to 3 will appear,
- A number less than or equal to one will appear,
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P(A) = 0.5, P(B) = 0.7, P(A ∩ B) = 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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| 1. | Harish | M | 30 |
| 2. | Rohan | M | 33 |
| 3. | Sheetal | F | 46 |
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| 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 |
| (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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