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Question
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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Solution
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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RELATED QUESTIONS
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.
Fill in the blank in following table:
| P(A) | P(B) | P(A ∩ B) | P(A ∪ B) |
| `1/3` | `1/5` | `1/15` | .... |
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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 |
| (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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