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Probability question

2024 · 31 Jan · Shift 1 · Q41
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  5. /2024 · 31 Jan · Shift 1 · Q41

Probability question

2024 · 31 Jan · Shift 1 · Q41

JEE MainMathematicsProbabilityMCQ+4 / −1
Two marbles are drawn in succession from a box containing 10 red, 30 white, 20 blue and 15 orange marbles, with replacement being made after each drawing. Then the probability, that first drawn marble is red and second drawn marble is white, is
  1. A
    425\frac{4}{25}254​
  2. B
    23\frac{2}{3}32​
  3. C
    225\frac{2}{25}252​
  4. D
    475\frac{4}{75}754​
View written solutionFree

Correct answer: D

  1. Total number of marbles

    The box contains: 10 red+30 white+20 blue+15 orange=7510\text{ red} + 30\text{ white} + 20\text{ blue} + 15\text{ orange} = 7510 red+30 white+20 blue+15 orange=75 So, total marbles =75= 75=75.

  2. Probability that the first marble is red

    Number of red marbles =10= 10=10. Hence, P(first is red)=1075=215P(\text{first is red}) = \frac{10}{75} = \frac{2}{15}P(first is red)=7510​=152​

  3. Probability that the second marble is white

    Since the marble is replaced, the total number of marbles remains 757575 for the second draw.

    Number of white marbles =30= 30=30. Therefore, P(second is white)=3075=25P(\text{second is white}) = \frac{30}{75} = \frac{2}{5}P(second is white)=7530​=52​

  4. Probability of both events happening in succession

    Because replacement is made, the two draws are independent. So, P(first red and second white)=P(first red)×P(second white)P(\text{first red and second white}) = P(\text{first red}) \times P(\text{second white})P(first red and second white)=P(first red)×P(second white) =215×25=475= \frac{2}{15} \times \frac{2}{5} = \frac{4}{75}=152​×52​=754​

  5. Check options

    The value obtained is: 475\frac{4}{75}754​ This matches Option D.

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