Question

The probability is 0.4 that a child exposed to a certain contagious disease will catch it. If catching the disease is independent among children, the probability that the fifth child exposed to the disease will be the third to catch it is ?

Answer 1

The probability that the fifth child exposed to a contagious disease will be the third to catch it, given a 0.4 probability of infection and independent events, is calculated using the binomial formula resulting in a probability of 13.824%.

Step-by-step explanation:

The student's question pertains to a concept in probability and involves finding the probability that the fifth child exposed to a certain contagious disease will be the third to catch it, given that each child has a probability of 0.4 of catching the disease and that catching the disease is independent among children. To answer this, we recognize that it represents a scenario which can be evaluated using the binomial probability formula. This scenario is equivalent to having two children catching the disease among the first four (with probability 0.4 each) and the fifth child catching it (with probability 0.4).

We can calculate this probability using a combination C(4,2) for the first four children and the individual probabilities:

P = C(4,2) × (0.4)3 × (0.6)2,

where C(n,k) represents the number of combinations of n items taken k at a time. Plugging in the numbers, we get:

P = 6 × (0.4)3 × (0.6)2

P = 6 × 0.064 × 0.36

P = 0.13824

Therefore, the probability that the fifth child exposed will be the third to catch the disease is 0.13824, or 13.824%.

Answer 2

The probability that the fifth child exposed to the disease will be the third to catch it is 0.05184.

Step-by-step explanation:

The probability that the fifth child exposed to the disease will be the third to catch it can be calculated using the binomial probability formula. Since catching the disease is independent among children, the probability that a child catches the disease is 0.4, and the probability that a child does not catch the disease is 0.6.

Let's calculate the probability that the first four children do not catch the disease and the fifth child does catch it:

P(4 children do not catch the disease) = (0.6)4 = 0.1296

P(the fifth child catches the disease) = 0.4

To find the probability that the fifth child is the third to catch the disease, we multiply these probabilities:

P(fifth child is the third to catch the disease) = P(4 children do not catch the disease) * P(the fifth child catches the disease) = 0.1296 * 0.4 = 0.05184