Exercise 17.2 1

X  B (n = 5, p = 0.6)



a) Th  e easiest way to calculate the probabilities given by the formula is by using a calculator. We can use either the Function menu or the List menu.





Or:







So, the table would be:

b)

x

0

1

2

3

4

5

P(x)

32 3125

48 625

144 625

216 625

162 625

243 3125

0.4 0.35 0.3 0.25 0.2 0.15 0.1 0.05 0 0







c) i)

1

2

3

4

5

µ = np ⇒ µ = 5 × 0.6 = 3 σ = npq ⇒ σ = 5 × 0.6 × 0.4 = 1.2 ≈ 1.095





ii) Again, the easiest way would be by using a calculator.



Exercise 17.2



d)



e) Now, we need to calculate the following probabilities: P (2  x  4) and P (1  x  5) .

Whilst the first one can be calculated directly, the second would be easier to calculate by using the complementary event.

Notice that the index of the list member is one more than the value of the random variable.





2

Let X be the number of respondents in favour of the decision. X  B (n = 20, p = 0.6) .



 20  5 a) P( X = 5) =  0.6 × 0.415 ≈ 0.001 294 4935   5 



 20  b) P( X = 0) =  0.60 × 0.4 20 ≈ 0.000 000 010 995  0  c) This problem is easier to solve by using the complementary event; therefore,



 e notice that the probability within one standard deviation, 0.8352, is much higher than the W empirical one, 0.6827. The probability within two standard deviations, 0.989 76, is fairly close to the empirical one, 0.9545.

P ( X  1) = 1 − P( X = 0) = 1 − 0.000 000 010 995 = 0.999 999 909 005.

d) E  ven though we can use the direct form, it is again easier to calculate the probability by using the complementary event; therefore,

e)

P ( X  2) = 1 − P ( X  1) = 1 − P ( X = 0) − P ( X = 1) = 1 − 0.000 000 010 995 − 0.000 000 329 85 = 0.999 999 6592.

µ = np ⇒ µ = 20 × 0.6 = 12 σ = npq ⇒ σ = 20 × 0.6 × 0.4 = 4.8 ≈ 2.19

3

a) In this part, we are going to use the binomial cumulative distribution function on a calculator.



Exercise 17.2



b) Explain it, if needed

Find the required probability

P ( x  3)

P( x  3)

0.929 53

3, 4, 5, 6

P ( x  3)

1 − P ( x  2)

0.255 69

More than 3

4, 5, 6

P ( x > 3)

1 − P ( x  3)

0.070 47

Fewer than 3

0, 1, 2

P ( x  2)

P( x  2)

0.744 31

Between 3 and 5 (inclusive)

3, 4, 5

P (3  x  5)

P ( x  5) − P ( x  2)

0.254 961

Exactly 3

3

P ( x = 3)

P( x = 3)

0.185 22

Number of successes x

List the values of x Write the probability statement

At most 3

0, 1, 2, 3

At least 3

4

a)











b)

I n the second GDC screen, we can see the cumulative probability values are all given except that for x = 7, but that one (which is the last term) is equal to 1. Explain it, if needed

Find the required probability

P ( x  3)

P ( x  3)

0.710 208

3, 4, 5, 6, 7

P ( x  3)

1 − P ( x  2)

0.580 096

More than 3

4, 5, 6, 7

P ( x > 3)

1 − P ( x  3)

0.289 792

Fewer than 3

0, 1, 2

P ( x  2)

P ( x  2)

0.419 904

Between 3 and 5 (inclusive)

3, 4, 5

P (3  x  5)

P ( x  5) − P ( x  2)

0.561 2544

Exactly 3

3

P ( x = 3)

P ( x = 3)

0.290 304

Number of successes x

List the values of x Write the probability statement

At most 3

0, 1, 2, 3

At least 3

5

a) This is not a binomial distribution since we don’t have a sequence of several independent trials with equal probabilities.



b) I f we choose the balls with replacement, then the trials become independent, so a sequence of three such trials with equal probabilities is a binomial distribution.

Exercise 17.2



5 c) Y  B  3,   8





6

7

 3   8

3− y

, y = 0, 1, 2, 3

y

0

1

2

3

P (Y = y )

0.0527

0.2637

0.4395

0.2441

d) It is easier to calculate the complementary event, which is all three green balls.



 3   5 y P (Y = y ) =      y   8

P ( y  2) = 1 − P ( y = 3) = 1 − 0.2441 = 0.7559

5 15 = = 1.875 8 8 5 3 45 f) V (Y ) = npq ⇒ V (Y ) = 3 × × = = 0.703 125 8 8 64 g) The complementary event of some green balls would be no green balls will be chosen; therefore, we calculate the probability: P ( y  1) = 1 − P ( y = 0) = 1 − 0.0527 = 0.9473. e) E (Y ) = np ⇒ E (Y ) = 3 ×

Since Nick guesses every single question, the probability that he chooses the correct answer from five 1 1 possible answers per question is ; therefore, the distribution is X  B 10,  .  5 5 a) b)–c)



1 d) E ( X ) = np ⇒ E ( X ) = 10 × = 2 5 There are 10 houses and in each we have an alarm system that is 98% reliable, so the distribution is X  B (10, 0.98) .

a) P ( X = 10)

b)

P ( X  5) = 1 − P ( X  4)

c) P ( X  8)

8

Notice that the result in part b can be interpreted as 1. Let X be the number of readers over 30 years of age. X  B (15, 0.4) .



a)–b)











c)



Exercise 17.2



We define the events by the letters A, B and C. Notice that the events satisfy the following relationship: A ∩ C = B, A ∪ C = U, where U is the universal set. Now, by using the addition formula for two sets, we calculate the probability of their union: P ( A ∪ C ) = P ( A) + P (C ) − P ( A ∩ C ) ⇒ 1 = P ( A) + P (C ) − P (B) .

9

Let X be the number of defective hard disks. X  B (50, 0.015) .



a) E ( X ) = np ⇒ E ( X ) = 50 × 0.015 = 0.75



 50  b) P( X = 3) =  0.0153 × 0.98547 ≈ 0.032 5112   3 



c) I t is easier to calculate the probability that more than one hard disk is defective by considering the complementary event, i.e. one or no hard disk is defective.

P ( X > 1) = 1 − P ( X  1) = 1 − P ( X = 0) − P ( X = 1) = 1 − 0.469 690 − 0.357 632 = 0.172 678 10 Let X be the number of ‘metallic grey’ cars. X  B (20, 0.1) .

a) P ( X  5) = 1 − P ( X  4) = 1 − 0.956 8255 = 0.043 1745 b) P ( X  6) = 0.997 614 c) P ( X > 5) = 1 − P ( X  5) = 0.011 2531



d) P (4  X  6) = P ( X  6) − P ( X  3) = 0.997 614 − 0.867 047 = 0.130 567



f) E ( X ) = np ⇒ E ( X ) = 100 × 0.1 = 10

e) If more than 15 are not ‘metallic grey’, then at least 4 are ‘metallic grey’. P ( X  4) = 0.956 826



g) σ = npq ⇒ σ = 100 × 0.1 × 0.9 = 9 = 3 h) A  ccording to the empirical rule, P ( µ − 2σ  X  µ + 2σ ) ≈ 0.95, and, in the probability model, we can calculate: 10 − 2 × 3  X  10 + 2 × 3 ⇒ 4  X  16 . So, a = 4 and b = 16.

11 Let X be the number of dogs with health insurance. X  B (100, 0.03) .

a) E ( X ) = np ⇒ E ( X ) = 100 × 0.03 = 3  100  b) P ( X = 5) =  0.0350.97 95 = 0.101 308  5  

c) P ( X  11) = 1 − P ( X  10) = 1 − 0.999 785 0751 = 0.000 214 925 12 Let X be the number of heads observed. X  B (5, 0.5) .

a)







b) From the table, we can read that P ( X = 0) = 0.031 25.



c) From the table, we can read that P ( X = 5) = 0.031 25.

Exercise 17.2



d) P ( X  1) = 1 − P ( X = 0) = 1 − 0.031 25 = 0.968 75



f) Since 2 heads are observed in every 10 tosses, the probability is 0.2.

e) If at least one tail is observed, then at most 4 heads are observed: P ( X  4) = 1 − P ( X = 5) = 1 − 0.031 25 = 0.968 75



From the table, we can read that P ( X = 0) = 0.032 768 .



From the table, we can read that P ( X = 5) = 0.000 32 .



P ( X  1) = 1 − P ( X = 0) = 1 − 0.327 68 = 0.672 32



If at least one tail is observed, then at most 4 heads are observed:

P ( X  4) = 1 − P ( X = 5) = 1 − 0.000 32 = 0.999 68 2 13 Let X be the number of hits observed. X  B  6,  .  5

 6 2 4 3 2 16 × 9 432 a) P ( X = 4) =       = 15 3 × 6 5 = 3125  4   5   5 5 b) John has to miss with his first two throws and then he needs to hit the target with his third throw.



3 2 2 18 P ( B) =   × =  5 5 125

Solution Paper 1 type

 2 14 Let X be the number of days Alice watched the news. X  B  5,  .  5



 5  2 4 3  5  2 5 16 3125 − 272 2853 P ( X  3) = 1 − (P ( X  4) + P ( X  5)) = 1 −        +      = 1 − (15 + 2) = = 3125   4   5   5  5   5   3125 3125 We used the complementary event since it has less elementary outcomes and is therefore easier to calculate.

Solution Paper 2 type

14 We are going to use the binomial cumulative distribution function.



Exercise 17.2

4 15 Let X be the number of cells that fail within a year. X  B 10,  .  5  10   4  10 410 a) P ( X = 10) =    = 10 ≈ 0.107 5  10   5  b) This is complementary to the event in part a. 410 P ( X  9) = 1 − P ( X = 10) = 1 − 10 ≈ 0.893 5 c) Again, we are going to use the complementary event since the calculation is much simpler. 4 n If we have n cells, the probability that all n cells fail is: P ( X = n) =   .  5 n 1 4 n 4 P ( X  n − 1) = 1 − P ( X = n) = 0.95 ⇒ 1 −   = 0.95 ⇒ =   ⇒  5 20  5  ln 20 ⇒ n (ln 4 − ln 5) = − ln 20 ⇒ n = ≈ 13.4 ln 5 − ln 4 So, we need at least 14 cells.