Math  /  Data & Statistics

QuestionQuestion 14 of 15, Step 1 of 1 Correct
A certain insecticide kills 70%70 \% of all insects in laboratory experiments. A sample of 11 insects is exposed to the insecticide in a particular experiment. What is the probability that exactly 2 insects will survive? Round your answer to four decimal places. Tables Keypad Answer

Studdy Solution

STEP 1

1. The insecticide kills 70% of insects, implying that the probability of an insect being killed is p=0.7p = 0.7.
2. Consequently, the probability of an insect surviving is q=1p=0.3q = 1 - p = 0.3.
3. The problem involves a fixed number of trials (11 insects), each with two possible outcomes (survive or not survive), making it a binomial distribution problem.
4. We need to find the probability that exactly 2 out of 11 insects will survive.

STEP 2

1. Define the binomial distribution parameters.
2. Use the binomial probability formula to calculate the probability of exactly 2 survivors.
3. Compute the binomial coefficient and the individual probabilities.
4. Multiply the binomial coefficient by the probabilities and round the result to four decimal places.

STEP 3

Define the parameters for the binomial distribution. Here, n=11n = 11 (number of trials), k=2k = 2 (number of successes, i.e., survivors), p=0.7p = 0.7 (probability of success, i.e., being killed), and q=0.3q = 0.3 (probability of failure, i.e., surviving).
n=11, k=2, p=0.7, q=0.3 n = 11, \ k = 2, \ p = 0.7, \ q = 0.3

STEP 4

Write the binomial probability formula:
P(X=k)=(nk)pnkqk P(X = k) = \binom{n}{k} p^{n-k} q^k Where (nk)\binom{n}{k} is the binomial coefficient.

STEP 5

Calculate the binomial coefficient (nk)\binom{n}{k}:
(112)=11!2!(112)!=11×102×1=55 \binom{11}{2} = \frac{11!}{2!(11-2)!} = \frac{11 \times 10}{2 \times 1} = 55

STEP 6

Compute the individual probabilities:
pnk=(0.7)112=(0.7)9 p^{n-k} = (0.7)^{11-2} = (0.7)^9 qk=(0.3)2 q^k = (0.3)^2

STEP 7

Calculate the numerical values of these probabilities:
(0.7)90.040353607 (0.7)^9 \approx 0.040353607 (0.3)2=0.09 (0.3)^2 = 0.09

STEP 8

Multiply the binomial coefficient by the probabilities:
P(X=2)=55×0.040353607×0.09 P(X = 2) = 55 \times 0.040353607 \times 0.09 P(X=2)55×0.00363182463 P(X = 2) \approx 55 \times 0.00363182463 P(X=2)0.19975035465 P(X = 2) \approx 0.19975035465

STEP 9

Round the result to four decimal places:
P(X=2)0.1998 P(X = 2) \approx 0.1998
Solution: The probability that exactly 2 insects will survive is approximately 0.19980.1998.

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