Counting the Wins in a Fixed Number of Goes

Mathematics · Probability · ages 16-17

Name ______________________ Date ____________

Every arrangement with the same number of successes has the same probability.
Circle one: True False
A fair coin is flipped 3 times. What is the probability of no heads at all?
Answer: ______________
A fair coin is tossed 4 times. What is P(exactly 2 heads)? Give your answer as a decimal.
Answer: ______________
Which of these is NOT one of the four conditions for a binomial model?
- the number of trials is fixed before you start counting them
- each trial has exactly two possible outcomes to choose between
- the number of successes is known in advance of the trials
- the probability of a success is the same on every trial
A checker tests 8 phones drawn without replacement from a box of 10 and treats the faulty count as binomial. Is this modelling choice sound?
Circle one: True False
A tombola is won on 0.1 of the tickets and somebody buys a strip of 6. Which of the four conditions is worth checking hardest?
- the fixed number of trials, since the strip of six tickets might be shared with a friend
- the two outcomes, since a single ticket could in theory be a spare
- the constant probability, since tickets already taken are not put back in the drum
- the number of successes, since nobody can know that before the strip is opened
A fair coin is flipped 4 times. What is the probability of 4 heads?
Answer: ______________
Which situation fits a binomial model?
- picking 4 pupils from a class of 30 and counting how many wear glasses
- rolling a die 20 times and counting how many rolls give a six
- measuring the heights of 20 pupils and recording each one
- drawing counters from a bag until the first red one appears
A quiz has 12 questions with 4 options each, answered entirely at random. Is the number of correct answers binomial?
- no, because guessing at random is not a real experiment
- no, because the four options mean that each question has four outcomes
- yes: 12 fixed questions, right or wrong, p = 0.25 each time, independent
- yes, but only if the person taking the quiz already knows some of the answers
If a binomial model does not fit, the arithmetic will produce an error rather than an answer.
Circle one: True False

Answer key

For grown-ups. Fold this page away before handing over the rest.

Counting the Wins in a Fixed Number of Goes W1-mt_X2kVGK67Md-s1

True · The same numbers are multiplied in a different order, and multiplication does not care about order. Three wins and seven misses gives the same product wherever the wins sit.
0.125 · Only one arrangement gives no heads, tails tails tails, and it has probability 0.5 × 0.5 × 0.5 = 0.125.
0.375 · There are 6 ways to choose which 2 of the 4 tosses land heads, out of 16 equally likely outcomes, so 6/16 = 0.375.
the number of successes is known in advance of the trials · The number of successes is the thing being modelled, so it cannot be known in advance. The other three are three of the four conditions, and independence is the fourth.
False · Without replacement the chance of a faulty phone changes with every draw, and drawing 8 from 10 changes it a lot, so independence fails.
the constant probability, since tickets already taken are not put back in the drum · Six tickets is fixed and a ticket wins or does not. Whether the drum still holds the same mix for the sixth ticket as for the first is the assumption that can fail.
0.0625 · Only one arrangement gives four heads, and its probability is 0.5 to the fourth power = 0.0625.
rolling a die 20 times and counting how many rolls give a six · Twenty rolls is fixed, each roll is a six or not, the chance stays at one sixth, and the rolls are independent. Picking pupils removes them, heights are not two outcomes, and drawing until red has no fixed number of trials.
yes: 12 fixed questions, right or wrong, p = 0.25 each time, independent · The outcome being counted is right or wrong, which is two outcomes whatever the number of options. Random guessing keeps p at 0.25 and leaves each question independent of the others.
False · The calculation runs perfectly happily on numbers that mean nothing. Only checking the four conditions tells you whether the answer describes anything.