I. In a certain species of spiders, the number of teeth on the chelicerae can be

Question

I. In a certain species of spiders, the number of teeth on the chelicerae can be 0, 1, 2 or 3. The probability that a randomly selected spider of that species will have one cheliceral tooth is 0.3; two cheliceral teeth is 0.4; and three cheliceral teeth is 0.25.

I-1. What is the probability that a randomly selected spider of the same species will have no cheliceral teeth?

I-2. What is the probability that a randomly selected spider of that species will have two or more cheliceral teeth?

I-3. If a scientist randomly selects two spiders of that species, what is the probability that both will have the same number of cheliceral teeth?

I-4. If a scientist randomly selects two spiders of that species, what is the probability that both will have different numbers of cheliceral teeth?

Explanation / Answer

Number of teeth on the chelicerae = {0, 1, 2, 3}

1) P(no teeth) = 1-(P(1 teeth)+P(2 teeth)+P(3 teeth))

P(no teeth) = 1-0.3-0.4-0.25 = 0.05

2) P(2 or more teeth)= P(2 teeth) + P(3 teeth)

P(2 or more teeth)= 0.4+0.25 = 0.65

3) P(both have equal teeth) = P(no teeth)*P(no teeth) + P(1 teeth)*P(1 teeth) + P(2 teeth)*P(2 teeth) + P(3 teeth)*P(3 teeth)

P(both have equal teeth) = 0.05*0.05 + 0.3*0.3 + 0.4*0.4 + 0.25*0.25

P(both have equal teeth) = 0.0025 + 0.09 + 0.16 + 0.0625 = 0.315

4) P(both have different teeth) = 1-P(both have equal teeth)

P(both have different teeth) = 1-0.315 = 0.685

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