{
  "problem": "Salim has a flock of $300$ chickens, $15\\%$ of which are infected with a virus. He plans on taking an SRS of $7$ chickens to test for the virus. Which of the following would find the probability that exactly $4$ of the $7$ chickens sampled have the virus? Choose 1 answer: Choose 1 answer: (Choice A) A ${7 \\choose 4}(0.15)^4(0.85)^3$ (Choice B) B ${7 \\choose 4}(0.15)^3(0.85)^4$ (Choice C) C ${300 \\choose 7}(0.15)^4(0.85)^3$ (Choice D) D $(0.15)^3(0.85)^4$ (Choice E) E $(0.15)^4(0.85)^3$",
  "hints": [
    "Probability of $4$ successes We want the probability that there are $4$ successes (infected chickens) in $7$ trials (number of chickens sampled), so we're going to need $3$ failures (not infected chickens) as well. The probability of each success is ${15\\%}$ and the probability of each failure is $85\\%}$. We can assume independence since we are sampling less than $10\\%$ of the population, so we can multiply probabilities to find the probability of getting $4$ successes followed by $3$ failures: $\\begin{aligned} P(\\text{SSSSFFF})&=({0.15})({0.15})({0.15})({0.15})(0.85})(0.85})(0.85}) \\\\\\\\ &=({0.15})^4(0.85})^3 \\end{aligned}$",
    "The binomial coefficient ${n \\choose k}$ SSSSFFF isn't the only arrangement that produces $4$ successes in $7$ trials. For instance, FFFSSSS would also produce the desired outcome. To count how many possible arrangements there are, we use the binomial coefficient ${n \\choose k}$. It tells us the number of possible arrangements for $k$ successes in $n$ trials. In this problem, we want $k=4$ successes (chicken is infected) in $n=7$ trials (number of chicken sampled), so we should use the binomial coefficient ${7 \\choose 4}$. [Tell me more about the binomial coefficient.]",
    "Putting it together Each arrangement has probability $(0.15)^4(0.85)^3$ so for our final answer we multiply this probability by the number of possible arrangements: ${7 \\choose 4}(0.15)^4(0.85)^3$",
    "The answer: ${7 \\choose 4}(0.15)^4(0.85)^3$"
  ]
}
{'source': 'AMPS', 'file': 'amps/khan/binomial-probability/16076776683.json'}