Download Statistics 1040 Review Exercises -

Statistics 1040
Review Exercises -- Chapter 14
More About Chance
Problem 1. Calculate the odds -- pair of dice
a. Chance both show 3 = chance first shows 3 AND second shows 3 = multiplication rule,
each die is independent, so simple multiplication rule applies:
P( both show 3) = (1 / 6) * ( 1 / 6) = 1/36
b. Chance both show SAME NUMBER = chance both show 1 OR both show 2 ... and so on
= (1/36) + (1/36) + (1/36) + (1/36) + (1/36) + (1/36)
= 6 / 36
Problem 2. Chance of moving exactly 11 spaces in Monopoly
= chance of rolling 11 = chance of getting 6 on first die AND 5 on second OR
chance of getting 5 on first die AND 6 on second
= (1/6) * (1/6) + (1/6) * (1/6)
= 2/36
See the grid of possible outcomes for two dice on p. 239
Problem 3. True or false and explain.
a. If die is rolled three times, probability of at least one ace is 1/6 + 1/6 + 1/6 = 3/6 = 1/2.
False. Double counts the possibility that you get two or three aces.
Use the complement rule to calculate the odds:
1 - P (no aces) = 1 - (pow 5/6 3) = 216 / 216 - 125 / 216 = 91 / 216
You could also calculate as:
P (an ace on any 1 throw) - P (at least 2 aces) = 1/2 - P (at least 2 aces)
but must then remember that there is a chance of getting 3 aces, so that
P (at least 2 aces) =
P (rolls 1 and 2 are aces) + P (rolls 2 and 3 are aces) + P (rolls 1 and 3 are aces) - P (all aces)
= 1 / 36 + 1 / 36 + 1/ 36 - 1 / 216 = 3 / 36 - 1 / 216 = 17 / 216
and having found that, solve the problem as:
1 / 2 - 17 / 216 = 108 / 216 - 17 / 216 = 91 / 216
You should better appreciate the complement rule if you try to work your way through this.
Problem 4. Which bet is better?
Two cards are dealt from a well-shuffled deck.
a. Bet 1: you get $ 1 if either of the first two cards is a queen.
= First card a queen OR second card a queen
= P (Q on 1) + P(Q on 2) - P (Q on both)
= 4 / 52 + 4 / 52 - (4 /52) * (3 / 51)
= 8 / 52 - 12 / (52 * 51)
= (408 - 12) / (52 * 51)
= 396 / 2652
= 0.1493 or 14.93 percent
b. Bet 2: get $ 1 if first card is a queen.
P (Q on 1) = 4 / 52 = 0.0769
Having more options to win always increases your odds of winning
Problem 5. Exclusive and independent.
These terms sound very much alike -- in fact, if one is the case, the other one is NOT.
a. If A and B are independent, P (A | B) = P (A)
But we define conditional probability as P (A | B) = P (A and B) / P (B)
Unless P (A) is zero, we CANNOT have P (A and B) = 0, that is, A and B can happen at the
same time, so A and B are NOT mutually exclusive.
b. If A and B are mutually exclusive, we know that P (A and B) = 0, so we know P (A | B) = 0.
Unless P (A) = 0, the two events CANNOT BE independent.
Problem 6. Multiplication and Addition rules.
a. Chance that at least one of the two events A and B will happen is equal to P (A) + P (B) if
A and B are MUTUALLY EXCLUSIVE.
b. Chance that both the events will happen is equal to P (A) * P(B) if both are INDEPENDENT.
Problem 7. Box model, with 4 draws with replacement from the box ( 1 2 2 3 3 )
Chance that 2 is drawn at least once will remain the same on each draw = 2 / 5
The chance 2 is NOT drawn is therefore 3/5 on each draw, so the complement rule tells us that
the chance it is drawn at least once is:
1 - (pow 3/5 4) = 81 / 625 = 0.1296
Problem 8. Box model, with 4 draws WITHOUT replacement from the box (1 2 2 3 3)
What are the chances 2 is drawn at least once. Simple – 100 percent.
Even if you draw a non-two on each of the first three draws, you will have to draw a
two the fourth time.
Problem 9. Two boxes: Box A = ( 1 2 3) Box B = ( 1 2 3 4)
Find the chance that:
a. Number drawn from A is larger than that drawn from B.
i.If you draw 1 from A, there is no chance of drawing a smaller number from B.
ii.If you draw 2 from A, you will draw a smaller number from B ¼ of the time.
You will draw 2 from A 1/3 of the time AND 1 from B ¼ of the time,
So the chance of this happening is 1/12.
iii.If you draw 3 from A, you will draw a smaller number from B ( 1 or 2) ½ the time.
The chance of this happening is 1/3 * ½ = 1/6.
The chance of (i), (ii) or (iii) happening is therefore 0 + 1/12 + 2/12 = 3/12 = ¼
b. The number drawn from A is equal to the one drawn from B.
If you draw 1 from A, you have a ¼ chance of matching it with your draw from B.
Chance of this is 1/3 * ¼ = 1/12.
Chance of drawing 2 from A AND 2 from B = 1/3 * ¼ = 1/12
Chance of drawing 3 form A AND 3 from B = 1/3 * ¼ = 1/12
Hence chance of match on 1, 2, OR 3 = 3 / 12.
c. The number drawn from A is SMALLER THAN the one drawn from B.
P( Draw 1 from A and 2, 3, or 4 from B) = 1/3 * ¾ = ¼ = 3 / 12
P (Draw 2 from A and 3 or 4 from B)
= 1/3 * ½ = 1/6 = 2 / 12
P (Draw 3 from A and 4 from B)
= 1/3 * ¼ = 1/12 = 1/ 12
So the chance that the draw from A is SMALLER is
3 / 12 + 2 /12 + 1 / 12 = 6 / 12 = ½
Problem 9 (continued).
You could also draw a box showing the possibilities. The symbols indicate the relation of A and B,
that is, = means A = B, < means A < B, and > means A > B
1
2
Draw from B
3
4
=
<
<
<
2
>
=
<
<
3
>
>
=
<
Draw from A
1
Problem 10. Which option is better?
a. Die rolled 60 times, you win $ 1 if ace or six, win $ 0 if anything else.
b. Sixty draws made at random with replacement from box ( 1 1 1 0 0 0), and
you are paid the amount shown in dollars.
Clearly, chance of winning on a draw from the box is 1/2, and in roll of die is 2/6 = 1/3.
Expected value of draw = 0.50, and of 100 draws = $ 30;
Expected value of roll = 0.3333 and of 60 rolls = $ 20
Problem 11. Deal three cards from well-shuffled deck.
a. chance that all are diamonds:
( 13 / 52) * (12 * 51) * (11 / 50) = 1716 / 132,600 = 0.0129
b. chance that none are diamonds:
( 39 / 52) * (38 / 51) * (37 / 50) = 54,834 / 132, 600 = 0.4135
c. Chance that NOT ALL are diamonds = chance that AT LEAST 1 is a non-diamond
Use complement rule = 1 – chance that ALL are diamonds
= 1 - ( 13 / 52) * (12 * 51) * (11 / 50) = 1 – 0.0129 = 0.9871
Problem 12. Coin is tossed 10 times.
a. Is it true that the chance of getting 10 heads in a row is 1 / 1024? YES
Since tosses are independent, the simple multiplication rule applies:
(pow ½ 10) = 1 / 1024
b. Is it true that if the first 9 tosses are heads, the chance of 10 heads in a row is ½ ? YES
Prob. (H | first 9 tosses heads) = ½ assuming that the coin is fair.
Problem 13. Box model with 2 red marbles and 98 blue ones. Draw with replacement.
How many draws are needed for there to be better than a 50 percent chance of getting a red marble?
Chance of at least one red marble in N draws = 1 - (pow 0.98 N)
Computer code:
(dotimes (i 50)
(writeln “~d ~A” (+ i 1) (- 1.0 (pow 0.98 (+ i 1)))
Partial results: 34 0.4969
35 0.5069
36 0.5168
So the answer will be 35 draws.
Problem 14. Lotto 6-53. You can usefully think about Lotto 2-53. The chance of any ticket winning if you only
have to match 2 numbers drawn without replacement is ( 1 / 53) * (1 / 52). If you have two tickets, the chances
are twice that. It does not matter whether the numbers are repeated on different tickets or not.