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Twelve students sit in a classroom, with seven students in the first row and the other five students in the second row. Three students are chosen randomly from the class.
The probability that exactly two of the three students chosen are in the first row is
\(B\)
\(\text{Pr(Exactly 2 from R1)}\ =\dfrac{\displaystyle \binom{7}{2}\binom{5}{1}}{\displaystyle \binom{12}{3}}=\dfrac{21}{44}\)
\(\Rightarrow B\)
A gym has 9 pieces of equipment: 5 treadmills and 4 rowing machines. On average, each treadmill is used 65% of the time and each rowing machine is used 40% of the time. --- 2 WORK AREA LINES (style=lined) --- --- 3 WORK AREA LINES (style=lined) --- i. \(P(T)=0.65, \ \ P(\overline{T})=1-0.65=0.35 \) \(P\text{(3 of 5 treadmills in use)}\ =\ ^5C_3 (0.65)^3(0.35)^2 \) ii. \(P(R)=0.4, \ \ P(\overline{R})=1-0.4=0.6 \) \(P\text{(no rowing machines in use)}\ =\ ^4C_0 (0.6)^4(0.4)^0=(0.6)^4 \) \(\text{Since 2 events are independent:} \) \(P\text{(3 treadmills and no rowing)}\ = \ ^5C_3 (0.65)^3(0.35)^2 \times (0.6)^4 \)
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The random variable `X` represents the number of successes in 10 independent Bernoulli trials. The probability of success is `p = 0.9` in each trial.
Let `r = P(X ≥ 1)`.
Which of the following describes the value of `r`?
`A`
`p = 0.9, \ \ barp = 0.1`
| `P(X >= 1)` | `= 1 – P(X = 0)` |
| `= 1 – (0.1)^10` | |
| `= 0.999…` |
`:. r > 0.9`
`=> A`
In a chocolate factory the material for making each chocolate is sent to a machines.
The time, `X` seconds, taken to produce a chocolate by machine is a binomial distribution where it can be shown that `P(X <= 3) = 9/32`.
A random sample of 10 chocolates is chosen. Find the probability, correct to two decimal places, that exactly 4 of these 10 chocolates took 3 or less seconds to produce. (2 marks)
`0.18`
`text(Let)\ \ C = \ text(number of chocolates that take less than 3 seconds)`
`C ∼\ text(Bin)(n, p) ∼\ text(Bin)(10, 9/32)`
| `P(C = 4)` | `=((10),(4)) (9/32)^4 (23/32)^6` |
| `=0.181…` | |
| `=0.18\ \ \ text{(2 d.p.)}` |
A school has a class set of 22 new laptops kept in a recharging trolley. Provided each laptop is correctly plugged into the trolley after use, its battery recharges.
On a particular day, a class of 22 students uses the laptops. All laptop batteries are fully charged at the start of the lesson. Each student uses and returns exactly one laptop. The probability that a student does not correctly plug their laptop into the trolley at the end of the lesson is 10%. The correctness of any student’s plugging-in is independent of any other student’s correctness.
Determine the probability that at least one of the laptops is not correctly plugged into the trolley at the end of the lesson. Give your answer correct to three decimal places. (2 marks)
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`0.902`
`text(Let)\ \ X = text(number not correctly plugged)`
`X\ ~\ text(Bin) (n,p)\ ~\ text(Bin) (22, 0.1)`
| `P(X>=1)` | `=1 – P(X=0)` |
| `=1-((n),(0)) (0.1^0)(0.9^22)` | |
| `=1 – 0.9^22` | |
| `=0.9015…` | |
| `=0.902\ \ text{(3 d.p.)}` |
It is known that 50% of the customers who enter a restaurant order a cup of coffee. If four customers enter the restaurant, what is the probability that more than two of these customers order coffee? (Assume that what any customer orders is independent of what any other customer orders.) (2 marks)
`5/16`
`X~\ text(Bin)(n,p)\ ~\ text(Bin)(4, 1/2)`
| `P(X > 2)` | `= P(X = 3) + P(X = 4)` |
| `= ((4),(3))(1/2)^3(1/2) + ((4),(4))(1/2)^4(1/2)^0` | |
| `= 4 xx 1/16 + 1 xx 1/16` | |
| `=5/16` |
A group of 12 people sets off on a trek. The probability that a person finishes the trek within 8 hours is 0.75.
Find an expression for the probability that at least 10 people from the group complete the trek within 8 hours. (2 marks)
`text(See Worked Solutions)`
`text{P(finishes within 8 hours)} = 0.75`
`text{P(not finish within 8 hours)} = 0.25`
`text(Let)\ \ X = text(number who finish below 8 hours)`
`:.\ text{P(at least 10 finish within 8 hours)}`
`=\ text{P(X=10) + P(X=11) + P(X=12)}`
`= ((12), (10)) (0.75)^10 (0.25)^2 + ((12), (11)) (0.75)^11 (0.25)^1 + ((12), (12)) (0.75)^12`
The probability that a particular type of seedling produces red flowers is `1/5`.
Eight of these seedlings are planted.
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i. `P(text{Red}) = 1/5,\ P(text{Not Red}) = 4/5`
`P(text(exactly 3 are red))`
`= \ ^8C_3 · (1/5)^3 · (4/5)^5`
ii. `P(text(none are red))`
`=\ ^8C_0 * (1/5)^0 * (4/5)^8`
`= (4/5)^8`
iii. `P(text(at least 1 is red))`
`= 1 – P(text(none are red))`
`= 1 – (4/5)^8`
In a large city, 10% of the population has green eyes.
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| i. | `P(text(G))` | `= 0.1` |
| `P(text(GG))` | `= 0.1 xx 0.1` | |
| `= 0.01` |
ii. `P(text(not G)) = 1 − 0.1 = 0.9`
`:. P(text(2 out of 20 have green eyes))`
`= \ ^(20)C_2 · (0.1)^2 · (0.9)^(18)`
`= 0.2851…`
`= 0.285\ \ \ text{(to 3 d.p.)}`
iii. `P(text(more than 2 have green eyes))`
`= 1 − [P(0) + P(1) + P(2)]`
`= 1 − [0.9^20 + \ ^20C_1(0.1)^1(0.9)^19 + \ ^20C_2(0.1)^2(0.9)^(18)]`
`= 1 − [0.1215… + 0.2701… + 0.2851…]`
`= 1 − 0.6769…`
`= 0.3230`
`= 0.32\ \ \ text{(to 2 d.p.)}`
The probability that it rains on any particular day during the 30 days of November is 0.1.
Write an expression for the probability that it rains on fewer than 3 days in November. (2 marks)
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`\ ^30 C_2 (0.1)^2 (0.9)^28 +\ ^30C_1 (0.1) (0.9)^29 + (0.9)^30`
| `P (R)` | `= 0.1` |
| `P (bar R)` | `= 1 – 0.1 = 0.9` |
`text(Over 30 days:)`
| `P (R<3)` |
`= P (R=2) + P(R=1) + P (R=0)` |
|
| `=\ ^30 C_2 (0.1)^2 (0.9)^28 +\ ^30C_1 (0.1)^1 (0.9)^29` | ||
| `qquad qquad +\ ^30C_0 (0.1)^0 (0.9)^30` | ||
| `=\ ^30C_2 (0.1)^2 (0.9)^28 +\ ^30C_1 (0.1)(0.9)^29 + (0.9)^30` |
A test consists of five multiple-choice questions. Each question has four alternative answers. For each question only one of the alternative answers is correct.
Huong randomly selects an answer to each of the five questions.
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i. `P text{(correct)} = 1/4`
`P text{(wrong)} = 3/4`
`P text{(3 correct, 2 wrong)}`
`=\ ^5C_3 * (1/4)^3 (3/4)^2`
`= (5!)/(3!2!) * (1/64) * (9/16)`
`= 90/1024`
`= 45/512`
ii. `P text{(3 or more correct)}`
`= P text{(3 correct)} + P text{(4 correct)} + P text{(5 correct)}`
`=\ ^5C_3 * (1/4)^3 (3/4)^2 +\ ^5C_4 * (1/4)^4 (3/4)^1 +\ ^5C_5 (1/4)^5 (3/4)^0`
`= 90/1024 + 15/1024 + 1/1024`
`= 53/512`
iii. `P text{(at least 1 incorrect)}`
`= 1\ – P text{(0 incorrect)}`
`= 1 -\ ^5C_5 (1/4)^5 (3/4)^0`
`= 1\ – 1/1024`
`= 1023/1024`
An examination has 10 multiple-choice questions, each with 4 options. In each question, only one option is correct. For each question a student chooses one option at random.
Write an expression for the probability that the student chooses the correct option for exactly 7 questions. (2 marks)
`\ ^10C_7 (1/4)^7 (3/4)^3`
| `P text{(Correct)}` | `= 1/4` |
| `P text{(Incorrect)}` | `= 3/4` |
`:. P text{(exactly 7 correct)} =\ ^10C_7 (1/4)^7 (3/4)^3`