Page 162 - NUMINO Challenge_D2
P. 162
2 Let A, B represent each of the two red cards Problem solving
and a represent the blue card. Then, there are 3 1 If the coin is flipped for the 4th and 5th times, the
outcomes for taking two cards out of the box,
which are (A, a), (B, a), and (A, B).Therefore, outcomes of having the number side or the
there is a higher possibility of having two cards
with different colors. If another blue card b is picture side are as shown below.
added to make the game fair, there are two
outcomes for having two cards with the same 4th 5th Matt wins
color, which are AB and ab, and there are four Matt wins
outcomes of having different colors, which are Number Side Number Side Matt wins
Aa, Ab, Ba, and Bb. If another red card C is Picture Side Sarah wins
added, there are three outcomes of having two Picture Side Number Side
cards with the same color, which are AB, AC, Picture Side
and BC, and there are also three outcomes of
having different colors, which are Aa, Ba, and There are three outcomes in which Matt wins and
Ca. Therefore, one red card should be added to
make the game fair. one outcome in which Sarah wins. Therefore from
the $12 that was bet, it is fair that
Matt gets $12 3 $9
4
1
and Sarah gets $12 4 $3.
Type 11-2 A Fair Sharing p.98~p.99 2 If the number side is marked with and the
picture side is marked with , the outcomes are
as shown below.
Winner
1 Third Fourth Fifth Sandra wins
Odd Odd Sandra wins
Odd Even Even Sandra wins
Odd Brian Wins
Even Odd Even Brian Wins
Odd Brian Wins
Even Even Brian Wins
Odd Brian Wins
Even The outcome of winning is four for both A and
B, therefore the game is fair.
2 In , there are 7 outcomes in which Sandra wins Creative Thinking p.100~p.101
and 1 outcome in which Brian wins.
3 There are 7 outcomes in which Sandra, wins so 1 If the hexagonal-shaped spinner
the probability is 7 There is one outcome in 2
8
1 is divided into six parts as shown 2 2
8
which Brian wins, so the probability is . on the right, there are two parts 10 4
Therefore, among the 16 marbles, it is fair that that have number 2, two parts 4
Sandra gets 16 7 14 marbles and Brian that have number 4, and one part that has
8
1 number 10.
gets 16 8 2 marbles.
(Average of the money that can be received)
200 3 400 2 1000 1
6 6 6
$ 6 $ 8 $ 10 $ 24 $4
6 6 6 6
Answer Key
and a represent the blue card. Then, there are 3 1 If the coin is flipped for the 4th and 5th times, the
outcomes for taking two cards out of the box,
which are (A, a), (B, a), and (A, B).Therefore, outcomes of having the number side or the
there is a higher possibility of having two cards
with different colors. If another blue card b is picture side are as shown below.
added to make the game fair, there are two
outcomes for having two cards with the same 4th 5th Matt wins
color, which are AB and ab, and there are four Matt wins
outcomes of having different colors, which are Number Side Number Side Matt wins
Aa, Ab, Ba, and Bb. If another red card C is Picture Side Sarah wins
added, there are three outcomes of having two Picture Side Number Side
cards with the same color, which are AB, AC, Picture Side
and BC, and there are also three outcomes of
having different colors, which are Aa, Ba, and There are three outcomes in which Matt wins and
Ca. Therefore, one red card should be added to
make the game fair. one outcome in which Sarah wins. Therefore from
the $12 that was bet, it is fair that
Matt gets $12 3 $9
4
1
and Sarah gets $12 4 $3.
Type 11-2 A Fair Sharing p.98~p.99 2 If the number side is marked with and the
picture side is marked with , the outcomes are
as shown below.
Winner
1 Third Fourth Fifth Sandra wins
Odd Odd Sandra wins
Odd Even Even Sandra wins
Odd Brian Wins
Even Odd Even Brian Wins
Odd Brian Wins
Even Even Brian Wins
Odd Brian Wins
Even The outcome of winning is four for both A and
B, therefore the game is fair.
2 In , there are 7 outcomes in which Sandra wins Creative Thinking p.100~p.101
and 1 outcome in which Brian wins.
3 There are 7 outcomes in which Sandra, wins so 1 If the hexagonal-shaped spinner
the probability is 7 There is one outcome in 2
8
1 is divided into six parts as shown 2 2
8
which Brian wins, so the probability is . on the right, there are two parts 10 4
Therefore, among the 16 marbles, it is fair that that have number 2, two parts 4
Sandra gets 16 7 14 marbles and Brian that have number 4, and one part that has
8
1 number 10.
gets 16 8 2 marbles.
(Average of the money that can be received)
200 3 400 2 1000 1
6 6 6
$ 6 $ 8 $ 10 $ 24 $4
6 6 6 6
Answer Key
