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Mathematics

In a family of 3 children, the probability of having at least one boy is:

  1. (18)\Big(\dfrac{1}{8}\Big)

  2. (58)\Big(\dfrac{5}{8}\Big)

  3. (34)\Big(\dfrac{3}{4}\Big)

  4. (78)\Big(\dfrac{7}{8}\Big)

Probability

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Answer

The sample space is: {BBB, BBG, BGB, GBB, BGG, GBG, GGB, GGG}

The total number of possible combinations = 8

Let E be the event of having at least one boy, then

E = {BBB, BBG, BGB, GBB, BGG, GBG, GGB}

The number of favorable outcomes to the event E = 7

∴ P(E) = Number of favorable outcomesTotal number of outcomes=78\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{7}{8}

Hence, option 4 is the correct option.

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