IB Mathematics SL 4.6 Use of Venn diagrams, tree diagrams AI HL Paper 1- Exam Style Questions- New Syllabus

(a)

The probabilities for the three morning activities must add to \(1\).

\(0.5+0.3+a=1\)

\(0.8+a=1\)

\(a=0.2\)

\(a=0.2\) A1

✅ Answer: \(0.2\)

(b)

To find the probability of ice-skating in the morning and climbing in the afternoon, multiply along the relevant branches of the tree diagram.

\(P(\text{ice-skating in morning})=0.5\)

\(P(\text{climbing in afternoon}\mid \text{ice-skating in morning})=0.3\)

So,

\(P(\text{ice-skating then climbing})=0.5\times0.3\) (A1)

\(P=0.15\)

\(P=0.15\), or \(15\%\). A1

✅ Answer: \(0.15\)

(c)

For go-karting in the morning and ice-skating in the afternoon, multiply the probability of choosing go-karting in the morning by the conditional probability of choosing ice-skating in the afternoon.

\(P(\text{go-karting in morning})=0.3\)

\(P(\text{ice-skating in afternoon}\mid \text{go-karting in morning})=b\)

Therefore,

\(P(\text{go-karting then ice-skating})=0.3b\) A1

✅ Answer: \(0.3b\)

(d)(i)

At the end of the day, \(87.5\%\) of students chose ice-skating. This includes students who chose ice-skating either in the morning or in the afternoon.

There are three relevant cases:

\(\text{ice-skating in morning}=0.5\)

\(\text{go-karting in morning and ice-skating in afternoon}=0.3b\)

\(\text{climbing in morning and ice-skating in afternoon}=0.2\times0.625\)

So,

\(0.5+0.3b+0.2(0.625)=0.875\) (M1)

\(0.5+0.3b+0.125=0.875\)

\(0.625+0.3b=0.875\)

\(0.3b=0.25\)

\(b=\dfrac{0.25}{0.3}\)

\(b=0.833333\ldots\)

\(b=0.833\), or exactly \(\dfrac{5}{6}\). A1

✅ Answer: \(b=0.833\), or \(\dfrac{5}{6}\)

(d)(ii)

After a student chooses go-karting in the morning, the two possible afternoon activities are ice-skating and climbing.

Therefore, their probabilities must add to \(1\):

\(b+c=1\)

\(c=1-b\)

Using \(b=\dfrac{5}{6}\),

\(c=1-\dfrac{5}{6}\)

\(c=\dfrac{1}{6}\)

\(c=0.166666\ldots\)

\(c=0.167\). A1

✅ Answer: \(c=0.167\), or \(\dfrac{1}{6}\)