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A room has three lightbulbs. Each one has a 30% probability of burning out within the month. Write the probability as a percentage rounded to one decimal place.

What is the probability that at the end of the month at least one of the bulbs will be lit?

Answer :

Answer:

97.3%

Step-by-step explanation:

Let the three bulbs be A, B and C respectively.

Let P(A) denote the probability that the first bulb will burn out

Let P(B) denote the probability that the second bulb will burn out

Let P(C) denote the probability that the third bulb will burn out

Now, we are told that Each one has a 30% probability of burning out within the month.

Thus;

P(A) = P(B) = P(C) = 30% = 0.3

Now, probability that at the end of the month at least one of the bulbs will be lit will be given as;

P(at least one bulb will be lit) = 1 - (P(A) × P(B) × P(C))

P(at least one bulb will be lit) = 1 - (0.3 × 0.3 × 0.3) = 0.973 = 97.3%

The probability that at the end of the month at least one of the bulbs will be lit is 97.3%.

Probability :

Let the three bulbs be [tex]E_{1},E_{2}[/tex] and [tex]E_{3}[/tex] respectively.

In three light bulbs Each one has a 30% probability of burning out within the month.

So that, [tex]P(E_{1}) = P(E_{2}) = P(E_{3}) = 0.3[/tex]

Now, probability that at the end of the month at least one of the bulbs will be lit,

[tex]P(E) = 1 - (P(E_{1}) *P(E_{2}) *P(E_{3}))\\\\P(E)=1-(0.3*0.3*0.3)\\\\P(E)=1-0.027=0.973[/tex]

Learn more about the probability here:

https://brainly.com/question/25870256

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