Reasoning · Grade 5-2 Counting Cases

Problem

Fair games and comparing likelihoods

Two experiments each have three possible outcomes. In (1) one die is rolled: an odd number, a number greater than 4, or a divisor of 6. In (2) two coins are tossed: two heads, exactly one head, or none. Pick the most likely outcome in each.
Your answer
How to solve
Strategy Make a Systematic List — When every result of an experiment is equally likely, comparing chances is nothing more than comparing counts, so the safe method is to write out every possible result once and then tick which ones belong to each case. For the die that list is short and obvious. For the two coins the list is the whole difficulty: it is tempting to write only 'two heads, one head, no heads' and think there are three equal results, but those three are not equally likely. Re-organising the list so that the two coins are told apart fixes that, and tossing a marked pair of coins a few times shows why it has to be done that way.
1STEP 1

Decide what 'most likely' will mean here

Read 'most likely' as the biggest count.

2STEP 2

Part (1): write out the six faces once

Write out the six faces.

{1, 2, 3, 4, 5, 6}
3STEP 3

Part (1): count each case

The three counts are 3, 2 and 4.

(i) 3, (ii) 2, (iii) 4
4STEP 4

Part (1): compare the counts

The largest is a divisor of 6.

3/6 = 1/2, 2/6 = 1/3, 4/6 = 2/3
5STEP 5

Part (2): the trap, and how to avoid it

Two coins must be told apart when counting.

6STEP 6

Part (2): list all four results and count

Write out all four results.

2 × 2 = 4; (i) 1, (ii) 2, (iii) 1; 1 + 2 + 1 = 4
7STEP 7

Part (2): compare, and see why one head is special

Exactly one head happens two ways, the most.

1/4, 2/4 = 1/2, 1/4
8STEP 8

Check by actually tossing

Tossing for real gives the same answer.

Answer
a divisor of 6, exactly 1 head
4/6, 2/4
In part (1) the three counts 3, 2 and 4 are each between 0 and 6, as counts of faces must be, and the winning chance of 4 out of 6 is a bit more than a half, which sounds right for a case that catches four of the six faces. In part (2) the counts 1, 2 and 1 add to exactly 4, the total number of results, so the three cases together cover everything that can happen and nothing has been counted twice — that check is what proves the list is complete. The answer to (2) also passes the common-sense test that two coins hardly ever agree with each other: they match only half the time, and matching splits into two heads or two tails, so each of those must be rarer than the mixed result.
Takeaway

When every result is equally likely, the most likely case is just the one you can list the most ways of getting - and two coins give four ways, not three!

  • Decide what 'most likely' will mean here
  • Part (1): write out the six faces once
  • Part (1): count each case
  • Part (1): compare the counts
  • Part (2): the trap, and how to avoid it
  • Part (2): list all four results and count
  • Part (2): compare, and see why one head is special
  • Check by actually tossing
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