MDM4UPage 1

Chapter5: Answer Section

1.a) There are sixteen possible outcomes.

1 + 1 = 2 / 2 + 1 = 3 / 3 + 1 = 4 / 4 + 1 = 5
1 + 2 = 3 / 2 + 2 = 4 / 3 + 2 = 5 / 4 + 2 = 6
1 + 3 = 4 / 2 + 3 = 5 / 3 + 3 = 6 / 4 + 3 = 7
1 + 4 = 5 / 2 + 4 = 6 / 3 + 4 = 7 / 4 + 4 = 8

b)

X / 2 / 3 / 4 / 5 / 6 / 7 / 8
P(X) / / / / / / /

c)E(X) = 2 + 3 + 4 + 5 + 6 + 7 + 8 = 5

2.a) Let X represent how many numbers were matched.

Based on these probabilities, we would expect that 5 816 302 tickets will match no numbers, 5 509 479 will match one number, and 1 766 235 will match two numbers. Combined with the data in the chart, this accounts for 13 322 590 tickets. The discrepancy is 17 553 tickets. This is a very small number in comparison to the size of the sample. The probability values are long-term averages. Any single draw may vary from the expected (predicted) percentages.

b)The experimental probability of matching 3, 4, 5 or 6 numbers is simply the frequency divided by the number of tickets.

3 / 0.0164
4 / 0.00085
5 / 0.0000185
6 / 0.000000075

3.

0.50 / 1.00 / 2.00 / Prize
H / H / H / 3.50
H / H / T / 1.50
H / T / H / 2.50
H / T / T / 0.00
T / H / H / 3.00
T / H / T / 0.00
T / T / H / 0.00
T / T / T / 0.00

All eight possible outcomes are equally likely. The total winnings for all 8 outcomes is $10.50. The expected value is $1.3125 per toss.

b)A fair game pays out as much as it takes in. The price per play should be the same as the expected value, which is (to the nearest cent) $1.31 per play.

4.a)

E(X) = 0 + 1 + 2 =

This is intuitively satisfying. Three drives is one-third of the inventory. We would expect to get one-third of the defective drives as a long-term average.

b)If we take the nine numbered cards from one suit, we could let two cards (say the 5 and 10) represent the defective drives. We could repeatedly draw three cards at random from the nine cards available and record the number of times the 5 and/or 10 were observed in the cards drawn. For ease of comparison, the simulation could be repeated 84 times and the frequencies for each possible outcome compared to the numerators of the theoretical probabilities.

5.(3a – 5b)6 =

6.The general term in the expansion is = .

We require the exponent to have a value of 6, so 36 – 5r = 6, which gives r = 6.

The required term is .

7. The smallest number of possible routes is 35 from A. The largest number of possible routes is 103 from C.

1 / 1
1 / 1 1
1 1 / 1 2 1
2 1 / 1 3 3 1
2 3 1 / 1 4 6 4
5 4 1 / 5 10 10 4
5 9 5 1 / 5 15 20 14
14 14 6 1 / 20 35 34 14

8.Pascal’s Identity states that or .

9.In the 1st situation, the probability of each required event is 0.5. The probability of 4 consecutive independent events is simply the product of their individual probabilities, so the combined probability is 0.54 or 0.0625.

The 2nd situation is the probability of one particular outcome in a binomial distribution. In this case, we have 4 different opportunities for the head to occur. An intuitive approach suggests that we multiply our 1st answer by 4 to obtain 0.25. Alternatively, we use the general term of the binomial expansion for a binomial experiment with n = 4 and p = 0.5.

10.

We can see that the probability of the second question is almost double that of the first. In the first example, we have an expected value of 11 and are requesting the probability that we deviate from the expected value by 27% over 100 trials. In the second example, we have an expected value of 5.5. We are still requesting the probability of deviating by 27% but it is for a sample that is twice as small. In general, we expect that the more trials we undertake, the closer our outcomes will be to the expected value.

11.Mike has confused expected value with the concept of a probability distribution. His idea with the bag and the marbles was a good start. Mike should record the number of red marbles observed in a draw of 20 and then repeat the drawing many times. The possible values for X will be 0, 1, 2, 3. The relative frequency with which these values occur over his many trials will form an experimental probability that approximates the actual binomial distribution.

12.Converting from a discrete to a continuous variable, we are looking for the probability that 6.5 < X < 12.5 for a binomial experiment with p = 0.16 and n = 50.

np = 8 > 5 and nq = 42 > 5 so we may use the normal approximation to the binomial distribution. = np=8.

 = = =

We need to find the z-scores for the left and right boundaries of the area required.

and z =

From tables, we obtain the area to the left of z = 1.736 is 0.9588 and the area to the left of z = –0.579 is 0.2814. The area between the two values is 0.6774.

The probability of between 7 and 12 defective parts is about 68%.

13.Nadia’s expected score will be the sum of the expected values for each of the four different situations. Use E(X) = np for each situation.

Score = 16(1) + 6 + 3 + 5 = 21.25

14. Each package in the sample is part of a binomial experiment with p = 0.15 and n = 40.

The normal approximation may be applied as np and nq are greater than 5.

and  = =

We require that no more than 20% of 40 packages are under weight.

To convert to a continuous variable, we require P(X8.5).

This corresponds to an area of about 0.8655. We have an 86.6% probability that a given sample will pass the test.