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Two boxes of pens are placed on the counter of a stationery shop. The first box contains 5 black pens and 6 red pens, while the second box contains 6 black pens and 5 red pens. If one pen is randomly taken out from the first box and placed in the second box, what is the probability that a randomly selected pen from the second box will be a red pen?

Note: Round your answer to 3 decimal places. For example, if the answer is 0.5, write 0.500; if the answer is 0.5447, write 0.545; and if the answer is 0.7825, write 0.783.

Answer :

To solve this problem, we need to calculate the probability that a randomly selected pen from the second box will be a red pen, after one pen is randomly taken from the first box and placed in the second box.

Let's break this down step-by-step:

  1. Initial Configuration:

    • First box: 5 black pens and 6 red pens.
    • Second box: 6 black pens and 5 red pens.

  2. Probability of Picking a Pen from the First Box:

    • Total pens in the first box = 5 black + 6 red = 11 pens.

  3. Scenarios When Picking a Pen from the First Box: (a) Picking a black pen.

    • Probability of picking a black pen = [tex]\frac{5}{11}[/tex].
    • If a black pen is moved to the second box:
      • Second box now has 7 black pens and 5 red pens.
      • Total pens in the second box = 12 pens.
      • Probability of picking a red pen from the second box = [tex]\frac{5}{12}[/tex].

  4. Scenarios When Picking a Pen from the First Box: (b) Picking a red pen.

    • Probability of picking a red pen = [tex]\frac{6}{11}[/tex].
    • If a red pen is moved to the second box:
      • Second box now has 6 black pens and 6 red pens.
      • Total pens in the second box = 12 pens.
      • Probability of picking a red pen from the second box = [tex]\frac{6}{12} = \frac{1}{2}[/tex] or 0.500.

  5. Total Probability Calculation:

    • The overall probability of picking a red pen from the second box can be calculated using the law of total probability:

    [tex]P(\text{red pen in second box}) = \left( \frac{5}{11} \times \frac{5}{12} \right) + \left( \frac{6}{11} \times \frac{1}{2} \right)[/tex]

    • Calculating each part:

      • [tex]\frac{5}{11} \times \frac{5}{12} = \frac{25}{132}[/tex]
      • [tex]\frac{6}{11} \times \frac{1}{2} = \frac{6}{22} = \frac{66}{132}[/tex]

    • Adding these probabilities:

    [tex]P(\text{red pen in second box}) = \frac{25}{132} + \frac{66}{132} = \frac{91}{132} \approx 0.689[/tex]

  6. Conclusion:

    • Therefore, the probability that a randomly selected pen from the second box will be a red pen is approximately 0.689.

Thus, the final answer, rounded to three decimal places, is 0.689.

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Rewritten by : Brahmana