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Example Marbles Drawn: Probability- 2 Video Lecture | CSAT Preparation - UPSC
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FAQs on Example Marbles Drawn: Probability- 2 Video Lecture - CSAT Preparation - UPSC
| 1. What is the probability of drawing a red marble from a bag of 10 marbles, where 3 are red and 7 are blue? |  |
| 2. If I draw 2 marbles without replacement from a bag containing 5 red marbles and 3 blue marbles, what is the probability of getting 2 red marbles? | |
Ans. When drawing without replacement, the probability of the second event is affected by the outcome of the first event. To find the probability of getting 2 red marbles, we multiply the probability of drawing a red marble on the first draw (5/8) by the probability of drawing another red marble on the second draw, given that the first marble was red (4/7). This gives us a probability of (5/8) * (4/7) = 20/56 or approximately 0.357.
| 3. If I draw 3 marbles with replacement from a bag containing 4 red marbles and 6 blue marbles, what is the probability of getting at least one red marble? | |
Ans. When drawing with replacement, each draw is independent, meaning the probability of drawing a red marble remains the same for each draw. To find the probability of getting at least one red marble, we can use the complementary approach. The probability of not getting any red marbles on a single draw is 6/10. Since we are drawing 3 marbles, the probability of not getting any red marbles in all three draws is (6/10)^3. Therefore, the probability of getting at least one red marble is 1 - (6/10)^3 = 1 - 0.216 = 0.784.
| 4. In a bag of marbles, there are 5 red marbles, 4 green marbles, and 6 blue marbles. If I draw a marble at random, what is the probability of drawing a red or green marble? | |
Ans. To find the probability of drawing a red or green marble, we need to add the probabilities of these two events occurring separately. The probability of drawing a red marble is 5/15, and the probability of drawing a green marble is 4/15. Adding these probabilities together gives us a total probability of (5/15) + (4/15) = 9/15 or 0.6.
| 5. If I draw 2 marbles with replacement from a bag containing 8 marbles, including 3 red marbles and 5 blue marbles, what is the probability of drawing exactly one red marble? | |
Ans. When drawing with replacement, each draw is independent, so the probability of drawing a red marble remains the same for each draw. To find the probability of drawing exactly one red marble, we need to consider two scenarios: drawing a red marble on the first draw and a blue marble on the second draw, or drawing a blue marble on the first draw and a red marble on the second draw. The probability of each scenario is (3/8) * (5/8) = 15/64. Since there are two mutually exclusive scenarios, the total probability is 2 * (15/64) = 30/64 or approximately 0.469.
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