Home / IBDP Maths AI: Topic: SL 4.5: Concepts of trial, outcome, equally likely outcomes: IB style Questions SL Paper 1

IBDP Maths AI: Topic: SL 4.5: Concepts of trial, outcome, equally likely outcomes: IB style Questions SL Paper 1

Question 

Gustav plays a game in which he first tosses an unbiased coin and then rolls an unbiased six-sided die.

If the coin shows tails, the score on the die is Gustav’s final number of points.
If the coin shows heads, one is added to the score on the die for Gustav’s final number of points.

(a) Find the probability that Gustav’s final number of points is 7.

(b) Complete the following table:

Final number of points 1 2 3 4 5 6 7
Probability              

(c) Calculate the expected value of Gustav’s final number of points.

▶️ Answer/Explanation

(a) Probability of scoring 7:

To score 7, Gustav must get heads on the coin and roll a 6 on the die:

\[ P(7) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12} = 0.0833 \]


(b) Completing the probability table:

Each number from 1 to 6 can be obtained in two ways:

  • Final score \(x\) is obtained from rolling \(x\) with tails.
  • Final score \(x\) is obtained from rolling \(x-1\) with heads.

Thus, for \( x = 1 \) to \( 6 \):

\[ P(x) = \frac{1}{12} + \frac{1}{12} = \frac{2}{12} = \frac{1}{6} \]

The completed table:

Final number of points 1 2 3 4 5 6 7
Probability \(\frac{1}{12}\) \(\frac{1}{6}\) \(\frac{1}{6}\) \(\frac{1}{6}\) \(\frac{1}{6}\) \(\frac{1}{6}\) \(\frac{1}{12}\)

(c) Expected Value Calculation:

Using the expected value formula:

\[ E(X) = (1 \times \frac{1}{12}) + (2 \times \frac{1}{6}) + (3 \times \frac{1}{6}) + (4 \times \frac{1}{6}) + (5 \times \frac{1}{6}) + (6 \times \frac{1}{6}) + (7 \times \frac{1}{12}) \]

Calculating:

\[ E(X) = \frac{1}{12} + \frac{2}{6} + \frac{3}{6} + \frac{4}{6} + \frac{5}{6} + \frac{6}{6} + \frac{7}{12} \]

\[ E(X) = \frac{1}{12} + \frac{4}{12} + \frac{6}{12} + \frac{8}{12} + \frac{10}{12} + \frac{12}{12} + \frac{7}{12} \]

\[ E(X) = \frac{48}{12} = 4 \]

Conclusion: The expected value of Gustav’s final number of points is 4.

…………………………..Markscheme…………………………..

  • (a) Correct probability calculation for scoring 7.
  • (b) Proper completion of the probability table.
  • (c) Correct use of the expected value formula.

Question 10. [Maximum mark: 7]

A game is played where two unbiased dice are rolled and the score in the game is the greater of the two numbers shown. If the two numbers are the same, then the score in the game is the number shown on one of the dice. A diagram showing the possible outcomes is given below.

                       

Let T be the random variable “the score in a game”.

(a) Complete the table to show the probability distribution of T . [2]

t

1

2

3

4

5

6

P(T = t)

      

(b) Find the probability that

(i)a player scores at least 3 in a game.

(ii)  a player scores 6, given that they scored at least 3. [3]

()Find the expected score of a game. [2]

▶️Answer/Explanation

(a)

 (b) (i) \(\frac{32}{36}(\frac{8}{9},0.888888…, 88.9%)\)

(ii) use of conditional probability  \(\frac{11}{32}\) (0.34375,34.4%)

(c)\(\frac{1 \times 1+3\times 2+5\times 3+…+11\times 6}{36}= \frac{161}{36}(4.4477222…)\)

Question 8. [Maximum mark: 7]

A game is played where two unbiased dice are rolled and the score in the game is the greater of the two numbers shown. If the two numbers are the same, then the score in the game is the number shown on one of the dice. A diagram showing the possible outcomes is given below.

                       

Let T be the random variable “the score in a game”.

(a) Complete the table to show the probability distribution of T . [2]

t

1

2

3

4

5

6

P(T = t)

      

(b) Find the probability that

(i)a player scores at least 3 in a game.

(ii)  a player scores 6, given that they scored at least 3. [3]

()Find the expected score of a game. [2]

▶️Answer/Explanation

(a)

  1. Note: Award A1 if three to five probabilities are correct. (b) (i) \(\frac{32}{36}(\frac{8}{9},0.888888…, 88.9%)\) (ii) use of conditional probability  \(\frac{11}{32}\) (0.34375,34.4%) (c)\(\frac{1 \times 1+3\times 2+5\times 3+…+11\times 6}{36}= \frac{161}{36}(4.4477222…)\)
  2. e.g. denominator of 32

Question

 A polygraph test is used to determine whether people are telling the truth or not, but it is not
completely accurate. When a person tells the truth, they have a 20% chance of failing the test.
Each test outcome is independent of any previous test outcome.
10 people take a polygraph test and all 10 tell the truth.
(a) Calculate the expected number of people who will pass this polygraph test.
(b) Calculate the probability that exactly 4 people will fail this polygraph test.
(c) Determine the probability that fewer than 7 people will pass this polygraph test.

▶️Answer/Explanation

Ans:

(a) \((E(X)=)10 \times 0.8\)
8 (people)
(b) recognition of binomial probability
0.0881 (0.0880803…)
(c) 0.8 and 6 seen OR 0.2 and 3 seen
attempt to use binomial probability
0.121  (0.120873…)

Question

 A group of 130 applicants applied for admission into either the Arts programme or the Sciences
programme at a university. The outcomes of their applications are shown in the following table.

(a) Find the probability that a randomly chosen applicant from this group was accepted by
the university.
An applicant is chosen at random from this group. It is found that they were accepted into the
programme of their choice.
(b) Find the probability that the applicant applied for the Arts programme.
Two different applicants are chosen at random from the original group.
(c) Find the probability that both applicants applied to the Arts programme.

▶️Answer/Explanation

Ans:

(a) \((\frac{17+25}{13}=) \frac{42}{130}(\frac{21}{65}, 0.323076…)\)
(b) \((\frac{17}{17+25}=)\frac{17}{42} (0.404761…)\)
(c) \(\frac{41}{130} \times \frac{40}{129}\)
\(=\frac{1640}{16770} \approx 0.0978 (0.0977936…, \frac{164}{1677})\)

Question

]The ticket prices for a concert are shown in the following table.

Ticket TypePrice (in Australian dollar, $)
Adult15
Child10
Student12
  • A total of 600 tickets were sold.
  • The total amount of money from ticket sales was $7816.
  • There were twice as many adult tickets sold as child tickets

Let the number of adult tickets sold be x , the number of child tickets sold be y , and the
number of student tickets sold be z.
(a) Write down three equations that express the information given above.
(b) Find the number of each type of ticket sold.

▶️Answer/Explanation

Ans:

(a) x + y + z = 600
15x + 10y + 12z = 7816
x = 2y
(b) x = 308, y = 154, z = 138

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