**Probability & Ratios (Grades 8-9)**

Calculators are allowed for all questions unless indicated by the question. There are 5 questions for this topic. The difficulty increases with the question number.

**Question 1:**

Red and black sweets are distributed amongst 5 people, John has 26 sweets, Stephan has 5 sweets, Susan has 14 sweets, James has 5 sweets and Josh has 7 sweets. The ratio of black to red sweets in total is 10:9 respectively. John has 70% of the total black sweets.

**A:** To one decimal place, what percentage of the total red sweets does John have?** (6 marks)**

**B: **Susan eats all of her sweets. The total ratio of black to red sweets for Stephan, James and josh is now 10:7. How many red sweets do Stephan, James and Josh have altogether?** (3 marks)**

Answer to part A

Total amount of sweets = $57$ (1 mark)

Amount of black sweets is equal to $57 \times \frac{10}{19} = 30$ (1 mark)

John has 70% of the 30 black sweets. $30 \times 0.7 = 21$ (1 mark)

$26-21 = 5$ red sweets. (1 mark)

Total amount of red sweets is $57- 30 = 27$ (1 mark)

John has 5 out of the 27 sweets, so $\frac{5}{27} \times 100 = 18.5%$ (1 mark)

Answer to part B

John has 5 red sweets and 21 black sweets (from the previous question) $57-21-5 = 31$ sweets left. (1 mark)

After Susan eats all her sweets, there are now 17 sweets left. $31- 14 = 17$ sweets for Stephan, James and Josh. (1 mark)

Using the given ratio, $\frac{7}{17}$ are red. Therefore they have **7 red sweets altogether**. (1 mark)

**Question 2:**

The temperature in a town on Monday and Tuesday can either be hot, cold or freezing. There is no chance of a storm occurring.

The probability that it will be hot on a Monday is $\frac{8}{23}$.

Whenever it is cold or freezing on Monday, the probability that it will be cold on Tuesday is $\frac{5}{7}$ and the probability that it will be freezing on Tuesday is $\frac{1}{7}$

**A. **What is the probability that it will not be hot on Monday, and be either cold or freezing on Tuesday? Give your answer in decimal form to 2 decimal places. **(4 marks)**

On Wednesday, the probability that there will be a storm is $\frac{17}{26}$. Whenever there is a storm the probability that it will also be freezing is $\frac{5}{7}$.

On Wednesday, the probability that it will be hot is always half the probability that it will be cold.

The temperature on Wednesday can either be hot, cold or freezing.

**B.** What is the probability that it will be cold or freezing on a Monday, hot on a Tuesday, and hot on Wednesday with a storm? Give your answer in standard form to 1 decimal place. **(5 marks)**

Answer to part A

$1- \frac{8}{23}= \frac{15}{23}$ = chance of the weather being cold or freezing on a Monday. (1 mark)

$\frac{15}{23} \times \frac{5}{7} = \frac{75}{161}$= The probability that it will be cold or freezing on Monday, and cold on Tuesday. (1 mark)

$\frac{15}{23} \times \frac{1}{7} = \frac{15}{161}$= The probability that it will be cold or freezing on a Monday, and freezing on a Tuesday. (1 mark)

$\frac{75}{161} + \frac{15}{161} = \frac{90}{161} = 0.56$(1 mark)

Award 4 marks for any alternative method, including $\frac{6}{7} \times \frac{15}{23}$

Answer to part B

$\frac{15}{23} \times \frac{1}{7} = \frac{15}{161} =$The probability that it will be cold or freezing on a Monday, and hot on a Tuesday. (1 mark)

Let the probability that it will be cold on a Wednesday when there is a storm be X, therefore the probability that it is hot is ½ X.

The total probability of it being either hot or cold when there is a storm is $\frac{2}{7}$, so $1.5x = \frac{2}{7}$, solve for $x$, $x = 0.0190, \frac{1}{2}x = 0.0095$ (2 marks)

$\frac{15}{161} \times \frac{17}{26} \times 0.095 = 5.8\times10^{-3}$ (2 marks)

Award 2 marks for an alternative method to find the probability of it being hot on Wednesday with a storm. Allow 1 mark for answer given in the incorrect form.

**Question 3:**

Stacy, Keith and John are doing their end of unit exams.

The probability that Stacy will pass is 0.85

The probability that Keith will pass is 0.75

The probability that John will pass is 0.8

**A.** What is the probability that all three pass the exam? **(2 marks)**

**B**. What is the probability that all three fail the exam? **(2 marks)**

**C.** What is the probability that at least one of them fails the exam? **(2 marks)**

**D.** What is the probability that only one of them fails? **(2 marks)**

Answer to part A

0.85 x 0.75 x 0.8 = **0.51** (2 marks)

Answer to part B

$0.15 \times 0.25 \times 0.2 =\frac{3}{400}$(2 marks)

Answer to part C

1-0.51 = **0.49** (the probability that at least 1 fails can be any combination other than all three passing (0.51)) (2 marks)

Answer to part D

(0.15 x 0.75 x 0.8) + (0.85 x 0.75 x 0.2) + (0.85 x 0.25 x 0.8) = **0.3875** (2 marks)

**Question 4:**

Box A contains 4 red pens and 4 green pens.

Box B contains 2 red pens and 5 green pens.

A pen is taken from box A and placed into box B. Then, a pen is taken from box B and place into box A.

**A.** Find the probability that box A has more red pens than green pens after these two moves. **(6 marks)**

Answer to part A

There are 4 total combinations of which pens are moved in and out of both boxes.

Scenario 1: Red pen removed from A to B, red pen removed from B to A.

Scenario 2: Red pen removed from A to B, green pen removed from B to A.

Scenario 3: Green pen removed from A to B, red pen removed from B to A.

Scenario 4: Green pen removed from A to B, green pen removed from B to A.

Probability of scenario 1 occurring: $\frac{4}{7} \times \frac{3}{8} = \frac{3}{14}$. Results in the same number of red pens in box A, $\frac{4}{7}$_{ }(1 mark)

Probability of scenario 2 occurring: $\frac{4}{7} \times \frac{5}{8} = \frac{5}{14}$ Results in 1 less red pen in box A, $\frac{3}{8}$. (1 mark)

Probability of scenario 3 occurring: $\frac{3}{7} \times \frac{2}{8} = \frac{3}{28}$. Results in 1 more pen in box A, $\frac{5}{6}$. (1 mark)

Probability of scenario 4 occurring: $\frac{3}{7} \times \frac{6}{8} = \frac{9}{28}$. Results in the same number of red pens in box A, $\frac{4}{7}$. (1 mark)

In scenario 1, 3 and 4, there are more red pens in box A than green pens. Therefore $\frac{4}{7} + \frac{5}{6} + \frac{4}{7} = \frac{9}{14}$(2 marks)

**Question 5:**

James buys a pack of ice creams, they contain 3 strawberry, 4 chocolate and 5 vanilla ice creams.

He eats 2 ice creams, picking them at random. Find the probability that:

**A.** Both ice creams are the same flavour **(4 marks)**

**B. **At least one is strawberry flavoured **(6 marks)**

Answer to part A

The probability of picking a strawberry is $\frac{3}{12}$, the probability of picking 2 strawberry ice creams is $\frac{3}{12} \times \frac{2}{11} $(because there are only 11 ice creams left after the first pick and 1 strawberry is missing, so there are only 2 strawberry ice creams left out of a total 11 ice creams)= $\frac{1}{22}$. (1 mark)

The probability of picking chocolate is $\frac{1}{3}$, the probability of picking 2 chocolate ice creams is $\frac{4}{12} \times \frac{3}{11} = \frac{1}{11}$_{ }(1 mark)

The probability of picking vanilla is $\frac{5}{12}$ the probability of picking 2 vanilla ice creams is $\frac{5}{12} \times \frac{4}{11} = \frac{5}{33}$ (1 mark)

$\frac{1}{22} + \frac{1}{11} + \frac{5}{33} = \frac{19}{66}$(1 mark)

Answer to part B

To get at least 1 strawberry ice cream, you must consider every scenario that gives a strawberry ice cream.

Scenario 1: 2 strawberries = $\frac{3}{12} \times \frac{2}{11} = \frac{1}{22}$(1 mark)

Scenario 2: Strawberry then Chocolate = $\frac{3}{12} \times \frac{4}{11} = \frac{1}{11}$ (1 mark)

Scenario 3: Strawberry then Vanilla = $\frac{3}{12} \times \frac{5}{11} = \frac{5}{44}$(1 mark)

Scenario 4: Chocolate then Strawberry = $\frac{4}{12} \times \frac{3}{11} = \frac{1}{11}$ (1 mark)

Scenario 5: Vanilla then Strawberry = $\frac{5}{12} \times \frac{3}{11} = \frac{5}{44}$ (1 mark)

Sum of all scenarios = $\frac{5}{11}$** (1 mark)**