Malebo has a bag which contains 4 red balls, 10 green balls and an unknown number of yellow balls. The probability of randomly choosing a green ball from the bag is 25%.

QUESTION 6

6.1.1 Show that there are 40 balls in the bag.

Step 1: Set up the probability equation. Let $R$ be the number of red balls, $G$ be the number of green balls, and $Y$ be the number of yellow balls. Given: $R = 4$, $G = 10$. Let the unknown number of yellow balls be $y$. Total number of balls ($N$) = $R + G + y = 4 + 10 + y = 14 + y$. The probability of choosing a green ball is 25%, which is 0.25. $P(\text{Green}) = \frac{\text{Number of green balls}}{\text{Total number of balls}}$ $0.25 = \frac{10}{14 + y}$

Step 2: Solve for $y$. $0.25(14 + y) = 10$ $3.5 + 0.25y = 10$ $0.25y = 10 - 3.5$ $0.25y = 6.5$ $y = \frac{6.5}{0.25}$ $y = 26$ So, there are 26 yellow balls.

Step 3: Calculate the total number of balls. $N = 14 + y = 14 + 26 = \boxed{\text{40 balls}}$ This shows that there are 40 balls in the bag.

6.1.2 Draw a tree diagram to represent all possible ways in which the two balls could have been drawn from the bag. Show the probabilities associated with EACH branch as well as the outcomes.

Number of balls: Red (R) = 4, Green (G) = 10, Yellow (Y) = 26. Total = 40. The balls are drawn without replacement.

• First Draw:

  • Probability of drawing a Red ball: $P(R_1) = \frac{4}{40}$
  • Probability of drawing a Green ball: $P(G_1) = \frac{10}{40}$
  • Probability of drawing a Yellow ball: $P(Y_1) = \frac{26}{40}$

• Second Draw (after one ball is removed, 39 balls remaining):

  • If the first ball was Red ($R_1$):

    • $P(R_2|R_1) = \frac{3}{39}$ (Outcome: RR)
    • $P(G_2|R_1) = \frac{10}{39}$ (Outcome: RG)
    • $P(Y_2|R_1) = \frac{26}{39}$ (Outcome: RY)

  • If the first ball was Green ($G_1$):

    • $P(R_2|G_1) = \frac{4}{39}$ (Outcome: GR)
    • $P(G_2|G_1) = \frac{9}{39}$ (Outcome: GG)
    • $P(Y_2|G_1) = \frac{26}{39}$ (Outcome: GY)

  • If the first ball was Yellow ($Y_1$):

    • $P(R_2|Y_1) = \frac{4}{39}$ (Outcome: YR)
    • $P(G_2|Y_1) = \frac{10}{39}$ (Outcome: YG)
    • $P(Y_2|Y_1) = \frac{25}{39}$ (Outcome: YY)

Tree Diagram Representation:

                                 / R (3/39) --> RR
                 / R (4/40) ----- G (10/39) --> RG
                |                \ Y (26/39) --> RY
                |
Start -------- G (10/40) ----- R (4/39) --> GR
                |                \ G (9/39) --> GG
                |                 \ Y (26/39) --> GY
                |
                 \ Y (26/40) ----- R (4/39) --> YR
                                 \ G (10/39) --> YG
                                  \ Y (25/39) --> YY

6.1.3 Calculate the probability that the two balls drawn will have the same colour.

Step 1: Identify the outcomes for the same colour. The outcomes where both balls have the same colour are RR, GG, and YY.

Step 2: Calculate the probability for each outcome. $P(RR) = P(R_1) \times P(R_2|R_1) = \frac{4}{40} \times \frac{3}{39} = \frac{12}{1560}$ $P(GG) = P(G_1) \times P(G_2|G_1) = \frac{10}{40} \times \frac{9}{39} = \frac{90}{1560}$ $P(YY) = P(Y_1) \times P(Y_2|Y_1) = \frac{26}{40} \times \frac{25}{39} = \frac{650}{1560}$

Step 3: Sum the probabilities and simplify. $P(\text{Same Colour}) = P(RR) + P(GG) + P(YY)$ $P(\text{Same Colour}) = \frac{12}{1560} + \frac{90}{1560} + \frac{650}{1560}$ $P(\text{Same Colour}) = \frac{12 + 90 + 650}{1560} = \frac{752}{1560}$ $P(\text{Same Colour}) = \boxed{\frac{94}{195}}$

QUESTION 7

7.1 Draw a Venn diagram to represent the information above.

Let R = Rugby, V = Volleyball, C = Chess. Total learners = 80. Given information:

• $|R| = 52$
• $|V| = 42$
• $|C \text{ only}| = 5$
• $|R \cap V \text{ only}| = 14$ (Rugby and Volleyball but not Chess)
• $|R \cap C \text{ only}| = 12$ (Rugby and Chess but not Volleyball)
• $|V \cap C \text{ only}| = 15$ (Volleyball and Chess but not Rugby)
• $|R \cap V \cap C| = x$ (All 3 types of sport)
• $|\text{None}| = 3$ (Did not like any sport)

We can fill the regions of the Venn diagram:

• The intersection of all three sports: $x$
• The intersection of R and V only: 14
• The intersection of R and C only: 12
• The intersection of V and C only: 15
• Chess only: 5
• Outside all circles (None): 3

Now, calculate the "only" regions for R and V:

• Rugby only: $|R| - (|R \cap V \text{ only}| + |R \cap C \text{ only}| + |R \cap V \cap C|) = 52 - (14 + 12 + x) = 52 - 26 - x = 26 - x$
• Volleyball only: $|V| - (|R \cap V \text{ only}| + |V \cap C \text{ only}| + |R \cap V \cap C|) = 42 - (14 + 15 + x) = 42 - 29 - x = 13 - x$

Venn Diagram Regions:

• R only: $26 - x$
• V only: $13 - x$
• C only: 5
• R and V only: 14
• R and C only: 12
• V and C only: 15
• R and V and C: $x$
• None: 3

(A visual representation of the Venn diagram would show three overlapping circles labeled R, V, C, with these values placed in their respective regions.)

7.2 Show that $x = 8$.

Step 1: Sum all the distinct regions in the Venn diagram. The sum of all regions must equal the total number of learners, which is 80. $(26 - x) + (13 - x) + 5 + 14 + 12 + 15 + x + 3 = 80$

Step 2: Simplify the equation. Combine the constant terms: $26 + 13 + 5 + 14 + 12 + 15 + 3 = 88$ Combine the $x$ terms: $-x - x + x = -x$ So the equation becomes: $88 - x = 80$

Step 3: Solve for $x$. $x = 88 - 80$ $x = \boxed{8}$ This shows that $x = 8$.

7.3 How many learners like rugby only?

Step 1: Use the expression for "Rugby only" from 7.1 and the value of $x$ from 7.2. Rugby only = $26 - x$ Substitute $x = 8$: Rugby only = $26 - 8 = \boxed{18}$ There are 18 learners who like rugby only.

7.4 Calculate the probability that a learner, chosen randomly, likes at least TWO different types of sport.

Step 1: Identify the number of learners who like at least two different types of sport. "At least two different types of sport" includes learners who like exactly two sports and learners who like all three sports. These are the regions:

• R and V only = 14
• R and C only = 12
• V and C only = 15
• R and V and C ($x$) = 8 Total learners who like at least two sports = $14 + 12 + 15 + 8 = 49$.

Step 2: Calculate the probability. Total number of learners = 80. $P(\text{at least two sports}) = \frac{\text{Number of learners who like at least two sports}}{\text{Total number of learners}}$ $P(\text{at least two sports}) = \frac{49}{80}$ $P(\text{at least two sports}) = \boxed{\frac{49}{80}}$