Question

question 9
9.1 for any two events, a and b, it is given that p(a)=0.48 and p(b)=0.26.
determine:
9.1.1 p(a and b) if a and b are independent events
9.1.2 p(a or b) if a and b are mutually exclusive events
9.2 a survey was conducted among 130 grade 11 learners to establish which snack they prefer to eat while they watch television. the results were summarised in the venn diagram below. however, some information is missing.
9.2.1 if 29 learners prefer at least two types of snacks, calculate the value of x and y.
9.2.2 determine the probability that a learner who does not eat nuts will either have another snack or no snack while watching television.
9.3 a group of 200 tourists visited the same restaurant on two consecutive evenings. on both evenings, the tourists could either choose beef (b) or chicken (c) for their main meal. the manager observed that 35% of the tourists chose beef on the first evening and 70% of them chose chicken on the second evening.
9.3.1 draw a tree diagram to show the different choices of main meals made on the two evenings. show on your diagram the probabilities associated with each branch as well as all the possible outcomes of the choices.
9.3.2 calculate the number of tourists who chose the same main meal on both evenings.
9.3.3 show that more tourists opted not to change their choice of main meal during their two visits to the restaurant.

Explanation:

9.1.1

Step1: Recall formula for independent events

For independent events \(A\) and \(B\), \(P(A\cap B)=P(A)\times P(B)\).

Step2: Substitute given values

Given \(P(A) = 0.48\) and \(P(B)=0.26\), then \(P(A\cap B)=0.48\times0.26 = 0.1248\).

9.1.2

Step1: Recall formula for mutually - exclusive events

For mutually - exclusive events \(A\) and \(B\), \(P(A\cup B)=P(A)+P(B)\).

Step2: Substitute given values

Given \(P(A) = 0.48\) and \(P(B)=0.26\), then \(P(A\cup B)=0.48 + 0.26=0.74\).

9.2.1

Step1: Calculate \(x\)

The number of learners who prefer at least two types of snacks is the sum of the elements in the intersections. So \(5 + 10+6+x=29\).
\[x=29-(5 + 10+6)=29 - 21=8\]

Step2: Calculate the total number of elements in the circles

The sum of elements in the three circles is \(41+5 + 8+10+25+6+30=125\).

Step3: Calculate \(y\)

Since there are 130 learners in total, \(y=130 - 125 = 5\).

9.2.2

Step1: Calculate the number of learners who do not eat nuts

The number of learners who do not eat nuts is \(41+10+30 + y=41+10+30 + 5=86\).

Step2: Calculate the number of learners who do not eat nuts and have another snack or no snack

The number of learners who do not eat nuts and have another snack or no snack is \(41+10+30+5 = 86\).

Step3: Calculate the probability

The probability \(P=\frac{86}{130}=\frac{43}{65}\approx0.6615\).

9.3.1

The tree - diagram has two levels. At the first level, the branches are \(B\) (probability \(P(B)=0.35\)) and \(C\) (probability \(P(C)=1 - 0.35 = 0.65\)).
From the \(B\) branch, the second - level branches are \(B\) (probability \(P(B)=0.3\)) and \(C\) (probability \(P(C)=0.7\)).
From the \(C\) branch, the second - level branches are \(B\) (probability \(P(B)=0.3\)) and \(C\) (probability \(P(C)=0.7\)).
The possible outcomes are \(BB\), \(BC\), \(CB\), \(CC\).

9.3.2

Step1: Calculate the number of tourists who choose \(BB\)

The number of tourists who choose \(BB\) is \(200\times0.35\times0.3=21\).

Step2: Calculate the number of tourists who choose \(CC\)

The number of tourists who choose \(CC\) is \(200\times0.65\times0.7 = 91\).

Step3: Calculate the total number of tourists who choose the same main meal

The total number of tourists who choose the same main meal is \(21+91=112\).

9.3.3

Step1: Calculate the number of tourists who change their choice

The number of tourists who choose \(BC\) is \(200\times0.35\times0.7 = 49\).
The number of tourists who choose \(CB\) is \(200\times0.65\times0.3=39\).
The total number of tourists who change their choice is \(49 + 39=88\).

Step2: Compare the numbers

Since \(112>88\), more tourists opted not to change their choice of main meal during their two visits to the restaurant.

Answer:

9.1.1: \(0.1248\)
9.1.2: \(0.74\)
9.2.1: \(x = 8,y = 5\)
9.2.2: \(\frac{43}{65}\)
9.3.1: Tree - diagram with first - level branches \(B\) (probability \(0.35\)) and \(C\) (probability \(0.65\)), second - level branches from \(B\): \(B\) (probability \(0.3\)) and \(C\) (probability \(0.7\)), from \(C\): \(B\) (probability \(0.3\)) and \(C\) (probability \(0.7\)), outcomes \(BB\), \(BC\), \(CB\), \(CC\)
9.3.2: \(112\)
9.3.3: Shown by calculating number of non - changers (\(112\)) and changers (\(88\)) and comparing them.