There are three urns, urn 1 having 5 white and 3 black balls, urn II with 4 white and 4 black balls and urn III with 3 white and 5 black balls. A fair die is tossed and
i) urn I is chosen if dice shows up 1, 2 or 3,
ii) urn II is chosen if dice shows up 4 or 5,
iii) urn III is chosen if dice shows up 6.
2 balls are drawn at random from the chosen urn. What is the probability that one of the balls drawn is white and the other black?
To find the probability that one of the balls drawn is white and the other is black, we can use the law of total probability and consider the probabilities for each urn separately and then combine them.
Let's calculate the probability for each case:
Case i) Urn I is chosen if the die shows up 1, 2, or 3:
- Probability of choosing urn I: P(urn I) = P(die shows up 1, 2, or 3) = 3/6 = 1/2 (since there are 6 sides on a fair die)
- Probability of drawing one white and one black ball from urn I:
P(1 white and 1 black | urn I) = P(white from urn I) * P(black from urn I) + P(black from urn I) * P(white from urn I)
P(1 white and 1 black | urn I) = (5/8) * (3/8) + (3/8) * (5/8) = 30/64
Case ii) Urn II is chosen if the die shows up 4 or 5:
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For the following frequency table obtain the relative frequencies and the cumulative frequencies of the ‘more than’ type.
| Life (in hours)
499.5 – 649.5 | Frequency
5 |
| 649.5 – 799.5 | 20 |
| 799.5 – 949.5 | 40 |
| 949.5 – 1099.5 | 81 |
| 1099.5 – 1249.5 | 57 |
| 1249.5 – 1399.5 | 34 |
| 1399.5 – 1549.5 | 13 |
| Total | 250 |
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