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प्रश्न
A box contains 30 bolts and 40 nuts. Half of the bolts and half of the nuts are rusted. If two items are drawn at random, what is the probability that either both are rusted or both are bolts?
उत्तर
The numbers of bolts and nuts are 30 and 40, respectively.
Then the numbers of rusted bolts and rusted nuts are 15 and 20, respectively.
Total number of items = 30 + 40 = 70
Total number of rusted items = 15 + 20 = 35
Total number of ways of drawing two items = 70C2
Let R and B be the events in which both the items drawn are rusted items and bolts, respectively.
R and B are not mutually exclusive events, because there are 15 rusted bolts.
P(items are both rusted or both bolts) = P (R ∪ B)
= P(R) + P (B) -P (R ∩ B)
=\[\frac{^{35}{}{C}_2}{^{70}{}{C}_2} + \frac{^{30}{}{C}_2}{^{70}{}{C}_2} - \frac{^{15}{}{C}_2}{^{70}{}{C}_2} = \frac{1 \times 2}{70 \times 69}\left( \frac{35 \times 34}{1 \times 2} + \frac{30 \times 29}{1 \times 2} - \frac{15 \times 14}{1 \times 2} \right)\]
\[= \frac{1850}{70 \times 69} = \frac{185}{483}\]
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| (b) If E1 and E2 are the mutually exclusive and exhaustive events | (ii) (E1 – E2) ∪ (E1 ∩ E2) = E1 |
| (c) If E1 and E2 have common outcomes, then | (iii) E1 ∩ E2 = Φ, E1 ∪ E2 = S |
| (d) If E1 and E2 are two events such that E1 ⊂ E2 | (iv) E1 ∩ E2 = Φ |
