Question

cory is a birdwatcher. he estimates that 30% of the birds he sees are american robins, 20% are dark - eyed juncos, and 20% are song sparrows. he designs a simulation. let 0, 1, and 2 represent american robins. let 3 and 4 represent dark - eyed juncos. let 5 and 6 represent song sparrows. let 7, 8, and 9 represent other birds. the table shows the simulation results. simulation results 01611 26343 87408 08889 58822 49003 49116 98444 34845 67970 63777 71890 01595 30500 43938 91971 58647 39440 28893 51995 what is the probability that at least one of the next five birds he sees is a robin? a. 0.70 b. 0.45 c. 0.65 d. 0.75

Explanation:

Step1: Count total simulations

There are 20 five - digit numbers in the simulation results, so the total number of simulations $n = 20$.

Step2: Count simulations with no robins

A simulation has no robins if it contains no 0, 1, or 2. The numbers without 0, 1, 2 are: 49003 (has 0, not valid), 49116 (has 1, not valid), 98444, 34845, 67970, 63777, 71890 (has 1, not valid), 30500 (has 0, not valid), 43938, 91971 (has 1, not valid), 58647, 39440 (has 0, not valid), 28893 (has 2, not valid), 51995 (has 1, not valid). The valid numbers (no robins) are 98444, 34845, 67970, 63777, 43938, 58647. So the number of simulations with no robins $m=6$.

Step3: Calculate probability of no robins

The probability of no robins in a five - bird simulation from the simulation results is $P(\text{no robins})=\frac{m}{n}=\frac{6}{20}=0.3$.

Step4: Calculate probability of at least one robin

The probability of at least one robin is the complement of the probability of no robins. Using the formula $P(A)=1 - P(\text{not }A)$, we have $P(\text{at least one robin}) = 1-0.3=0.7$.

Answer:

A. 0.70