AMC 8 Daily Practice Round 9

Complete problem set with solutions and individual problem pages

Problem 15 Easy

A switch is connected to three independent light bulbs, meaning each bulb operates independently of the others. When the switch is pressed, each bulb has a 50\% probability of turning on. What is the probability that at least two bulbs will light up with a single press of the switch?

  • A.

    \frac13

  • B.

    \frac38

  • C.

    \frac12

  • D.

    \frac23

  • E.

    \frac56

Answer:C

The probability of two bulbs being on and one bulb being off is:   P_1 = \binom{3}{2} \left( \frac{1}{2} \right)^2 \times \frac{1}{2} = \frac{3}{8}

The probability of all three bulbs being on is:   P_2 = \left( \frac{1}{2} \right)^3 = \frac{1}{8}

Thus, the total probability is:   P = P_1 + P_2 = \frac{3}{8} + \frac{1}{8} = \frac{1}{2}

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