Lesson 4.1 Checkpoint

Suppose a married man and woman both carry a gene for cystic fibrosis but don’t have the disease themselves. According to the laws of genetics, the probability that any child they have will develop cystic fibrosis is 0.25.

Explain what this probability means.

(LT 4.1.1 #1)

• If you take a very large random sample of children whose parents both carry the gene for cystic fibrosis but don’t have the disease themselves, about 25% of the children will develop cystic fibrosis.

• If you take a very large random sample of children whose parents both carry the gene for cystic fibrosis but don’t have the disease themselves, about 0.25% of the children will develop cystic fibrosis.

• About 25% of children will develop cystic fibrosis.

• Exactly 0.25% of children will develop cystic fibrosis.

Suppose a married man and woman both carry a gene for cystic fibrosis but don’t have the disease themselves. According to the laws of genetics, the probability that any child they have will develop cystic fibrosis is 0.25.

If the couple has 4 children, is one of them guaranteed to get cystic fibrosis? Explain.

(LT 4.1.1 #2)

• No; probability describes what happens in many, many repetitions (way more than 4) of a chance process. We would expect to get about 1 child who develops cystic fibrosis in a random sample of 4 children, but this result is not guaranteed.

• Yes; probability describes what happens in many, many repetitions of a chance process. We would expect to get about 1 child who develops cystic fibrosis in a random sample of 4 children, but this result is not guaranteed.

• No; probability describes what happens in many, many repetitions (way more than 4) of a chance process. We would expect to get exactly 1 child who develops cystic fibrosis in a random sample of 4 children, and this result is guaranteed.

• Yes; probability describes what happens in many, many repetitions (way more than 4) of a chance process. We would expect to get exactly 1 child who develops cystic fibrosis in a random sample of 4 children, and this result is guaranteed.

A very good professional baseball player gets a hit about 35% of the time over an entire season. After the player failed to hit safely in six straight at-bats, a TV commentator said, “He is due for a hit by the law of averages.” Explain why the commentator is wrong.

(LT 4.1.2 #1)

• The commentator’s claim is based on the erroneous “law of averages.” Even after the player failed to hit safely in six straight at-bats, he will continue to have the same 35% chance of getting a hit on the next at-bat.

• The commentator’s claim is based on the erroneous “law of large numbers.” Even after the player failed to hit safely in six straight at-bats, he will continue to have the same 35% chance of getting a hit on the next at-bat.

• The commentator’s claim is based on the erroneous “law of averages.” Even after the player failed to hit safely in six straight at-bats, he will higher chance of getting a hit on the next at-bat.

• The commentator’s claim is based on the erroneous “law of large numbers.” Even after the player failed to hit safely in six straight at-bats, he will have a higher chance than 35% chance of getting a hit on the next at-bat.

Imagine tossing a coin 6 times and recording heads (H) or tails (T) on each toss. Which of the following outcomes is more likely: HTHTTH or TTTHHH? Justify your answer.

(LT 4.1.2 #2)

• They are equally likely, because any sequence of six tosses is 1/64.

• They are not equally likely, because any sequence of six tosses is 1/64.

• They are equally likely, because any sequence of six tosses is 1/2.

• They are not equally likely, because any sequence of six tosses is 1/2.

A randomly selected U.S. adult male has probability about 0.07 of having some form of red–green color blindness. Suppose we choose 4 U.S. adult males at random. What’s the probability that at least one of them is red–green color-blind? Design and carry out a simulation to help answer this question.

What is the State part to this simulation?

(LT 4.1.3 #1)

• What is the probability that at least 1 of 4 randomly selected adult males is red-green color-blind (assuming the probability of having some form of red-green color-blindness is 0.07)?

• Let 127 5 adult male has red- green color-blindness and 82100 5 adult male does not have red-green color-blindness. Use randInt(1,100,4) to simulate taking a random sample of 4 adult males. Record the number of adult males in the sample of 4 who have red- green color-blindness.

• 
• In 50 repetitions, there was at least 1 color-blind adult male 12 times. Assuming the probability of having some form of red-green color-blindness is 0.07, the estimated probability that at least 1 of 4 randomly selected adult males is red-green color-blind is approximately 12/50 5 24%.

A randomly selected U.S. adult male has probability about 0.07 of having some form of red–green color blindness. Suppose we choose 4 U.S. adult males at random. What’s the probability that at least one of them is red–green color-blind? Design and carry out a simulation to help answer this question.

What is the Conclude part to this simulation?

(LT 4.1.3 #2)

• What is the probability that at least 1 of 4 randomly selected adult males is red-green color-blind (assuming the probability of having some form of red-green color-blindness is 0.07)?

• Let 127 5 adult male has red- green color-blindness and 82100 5 adult male does not have red-green color-blindness. Use randInt(1,100,4) to simulate taking a random sample of 4 adult males. Record the number of adult males in the sample of 4 who have red- green color-blindness.

• 
• In 50 repetitions, there was at least 1 color-blind adult male 12 times. Assuming the probability of having some form of red-green color-blindness is 0.07, the estimated probability that at least 1 of 4 randomly selected adult males is red-green color-blind is approximately 12/50 5 24%.