Probability, Part 2 - Cascade Ridge Math Club

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Name: _________________________________
Compound Probability
March 12, 2012
1. The game of backgammon uses two standard dice, each with the numbers one through six. You
need to roll double 3s to win the game. What is the probability you will get that result on your
next roll?
2. Your drawer contains 6 red socks and 11 blue socks. It's too dark to see which are which, but
you grab two anyway. What is the probability that both socks are red?
3. Stephanie tossed a die onto a black-and-red checkerboard. What is the probability that it will
land with a value less than 3 and on a red square?
4. Suppose that a test has six questions, each of which can be answered either true or false. If
each question is answered randomly, what is the probability of correctly answering all six
questions? Express your answer as a reduced fraction.
5. Suppose that the probability that July 4th will be rainy is 1/10, the probability that July 3rd is
sunny is 3/4, and the probability that July 5th is sunny is also 3/4. What is the probability that all
three days will be rainy? Express your answer as a reduced fraction.
6. In the Seattle Mariners’ historical 2001 season, Edgar Martinez batted 0.306 (meaning 30.6% of
the times he was at the plate he got a hit), Ichiro Suzuki batted 0.350, and Bret Boone batted
0.331. If these three players were to each come up to bat one more time, what is the
probability that none of them would get a hit? Express your answer as a decimal rounded to the
nearest thousandth.
7. Suppose that the probability that you will drop a penny on the ground is 1/5, and the probability
that you will find a penny on the ground today is 1/4. If the two events are independent, what is
the probability that at least one of the two events will occur? Express your answer as a percent.
8. How many times should a coin be tossed to have at least a 95% chance of it landing “heads”
up at least once?
Cascade Ridge PTSA Math Club
1
Show your work on a separate sheet of paper.
Name: _________________________________
Compound Probability
March 12, 2012
BONUS PROBLEMS
9. An animal cage is holding 7 black cats and 5 white cats. None of them want to be in there. The
cage door is opened slightly and two cats escape. What is the probability that the escaping cats
are both white?
10. The names of 10 boys and 11 girls from your class are put into a hat. What is the probability that
the first two names chosen will be a boy followed by a girl?
11. What is the probability of rolling an even number with a fair six-sided die followed by flipping
heads twice in a row with a fair coin?
12. A drawer contains 18 socks, some black and some white. Two socks are randomly drawn from
the drawer. The probability that both socks are black is 26/51. How many socks in the drawer
are white?
Cascade Ridge PTSA Math Club
2
Show your work on a separate sheet of paper.
Name: _________________________________
Compound Probability
March 12, 2012
SOLUTIONS:
1. 1/36
We can consider the two dice to be two independent events. The probability of getting a 3
on the first die is 1/6. The probability of getting a 3 on the second die is also 1/6. Therefore,
1
1
1
× =
6 6 36
2. 15/136
We can consider drawing two socks from the drawer to be two separate events. When we pull
out the first sock, the probability of it begin red is 6/17. Then, when we pull out the second
sock, there are only 16 socks left in the drawer, 5 of which are red. Therefore, for the second
sock, the probability of getting a red is 5/16. So,
6
5
×
=
17 16
30
272
=
15
136
3. 1/6
The probability that the die will show a number less than three is 2/6, or 1/3. The probability
that it will land on a red square is 1/2. So,
2
6
1
2
×2 =
12
=
1
6
4. 1/64
The probability of answering any single question correctly is 1/2. There are six questions, and all
six questions can be considered as independent events. So,
1
1
1
1
1
1
1
× × × × × =
2 2 2 2 2 2 64
5. 1/160
The probabilities that each day individually will be rainy are:
July 3
1/4
July 4
1/10
July 5
1/4
So, the probability that all three days will be rainy is:
1
10
1
1
×4×4 =
1
160
6. 0.302
Cascade Ridge PTSA Math Club
3
Show your work on a separate sheet of paper.
Name: _________________________________
Compound Probability
March 12, 2012
The probabilities of each player individually not getting a hit are:
Edgar
Ichiro
Bret
1 – 0.306 = 0.694
1 – 0.350 = 0.650
1 – 0.331 = 0.669
So, the probability of all three players not getting a hit is:
0.694 × 0.650 × 0.669 = .3017859
Rounding this to the nearest thousandth, we get 0.302
7. 40%
The probability of dropping a penny plus the probability of finding a penny = 9/20. But then we
need to subtract the probability of both events occurring which is 1/5 × 1/4 = 1/20. So 9/20 –
1/20 = 8/20 = 4/10 = 40%. If we think of it like a Venn diagram:
8. 5
To say that there is at least a 95% chance of landing HEADS at least once is equivalent to saying
that there is less than 5% chance of it landing TAILS every time. So, the question becomes, how
many times would the coin have to be tossed for there to be less than 5% chance of it landing
TAILS every time.
If we toss the coin only once, then the probability of getting TAILS every time is:
1
2
= 0.5 = 50%
If we toss the coin twice, then the probability of getting TAILS every time is:
1
2
×
1
2
=
1
4
= 0.25 = 25%
Cascade Ridge PTSA Math Club
4
Show your work on a separate sheet of paper.
Name: _________________________________
Compound Probability
March 12, 2012
If we toss the coin three times, then the probability of getting TAILS every time is:
1
2
×
1
2
×
1
2
1
8
=
= 0.125 = 12.5%
If we toss the coin four times, then the probability of getting TAILS every time is:
1
2
×
1
2
×
1
2
×
1
2
=
1
16
= 0.0625 = 6.25%
If we toss the coin five times, then the probability of getting TAILS every time is:
1
2
×
1
2
×
1
2
×
1
2
×
1
2
=
1
32
= 0.03125 = 3.125%
We’re finally less than 5%. So, the answer is five times.
9. 5/33
5
·
4
12 11
20
5
110
11
= 132 = 33
10. 11/42
10 11
·
21 20
= 420 = 42
11. 1/8
1 1 1
1
· · =8
2 2 2
12. 5
b
18
·
b−1
17
b(b−1)
306
26
= 51
26
= 51
51b(b-1) = 7956
b(b-1) = 156
Since b must be 17 or less, and b-1 = the next lowest number, we try:
17 · 16 = 272
16 · 15 = 240
15 · 14 = 210
14 · 13 = 182
13 · 12 = 156  There must be 13 black socks
Cascade Ridge PTSA Math Club
5
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