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Two patients are randomly selected. The probability that a patient tests positive for malaria is 0.6. What is the probability that exactly one of the patients tests negative?

Mathematics

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Answers

2 parts
a
0.48
b
0.16

Step-by-step solution

5 steps

Let PP be the event that a patient tests positive for malaria, and NN be the event that a patient tests negative for malaria.

From the problem, we are given:

P(P)=0.6P(P) = 0.6

This can be written as a fraction:

P(P)=610P(P) = \frac{6}{10}

The probability that a patient tests negative is:

P(N)=1−P(P)=1−0.6=0.4P(N) = 1 - P(P) = 1 - 0.6 = 0.4

As a fraction:

P(N)=410P(N) = \frac{4}{10}

Part a· 3 steps

The probability that one patient tested negative while the other tested positive.

This means one patient is positive and the other is negative. Since the order is not specified, there are two possibilities: (Patient 1 positive AND Patient 2 negative) OR (Patient 1 negative AND Patient 2 positive). Since the selections are random, the events are independent.

  1. Identify the probabilities for positive and negative tests.

    P(P)=610P(P) = \frac{6}{10} P(N)=410P(N) = \frac{4}{10}

  2. Calculate the probability for each specific order.

    The probability of Patient 1 positive and Patient 2 negative is P(P)×P(N)P(P) \times P(N).
    The probability of Patient 1 negative and Patient 2 positive is P(N)×P(P)P(N) \times P(P).

  3. Sum the probabilities for the two possible orders.

    P(one negative, one positive)=(P(P)×P(N))+(P(N)×P(P))P(\text{one negative, one positive}) = \left(P(P) \times P(N)\right) + \left(P(N) \times P(P)\right) P(one negative, one positive)=(610×410)+(410×610)P(\text{one negative, one positive}) = \left(\frac{6}{10} \times \frac{4}{10}\right) + \left(\frac{4}{10} \times \frac{6}{10}\right) P(one negative, one positive)=24100+24100P(\text{one negative, one positive}) = \frac{24}{100} + \frac{24}{100} P(one negative, one positive)=48100P(\text{one negative, one positive}) = \frac{48}{100} P(one negative, one positive)=0.48P(\text{one negative, one positive}) = \boxed{0.48}

Part b0.16

The probability that both patients tested negative.

This means Patient 1 tested negative AND Patient 2 tested negative. Since the selections are random, the events are independent.

  1. Identify the probability for a negative test.

    P(N)=410P(N) = \frac{4}{10}

  2. Calculate the probability that both patients test negative.

    P(both negative)=P(N)×P(N)P(\text{both negative}) = P(N) \times P(N) P(both negative)=410×410P(\text{both negative}) = \frac{4}{10} \times \frac{4}{10} P(both negative)=16100P(\text{both negative}) = \frac{16}{100} P(both negative)=0.16P(\text{both negative}) = \boxed{0.16}
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