(a) The probability that Shalini is late for school on any day is \( \frac{1}{6}\)
(i) Complete the tree diagram for Monday and Tuesday.
(ii) Calculate the probability that Shalini is late on Monday but is not late on Tuesday.
(b) The Venn diagram shows the number of students in a group of 50 students who wear glasses (G), who wear trainers (T) and who have a mobile phone (M).
(i) Use set notation to describe the region that contains only one student.
(ii) Find \( n\left ( T’\cap (G\cup M) \right )\)
(iii) One student is picked at random from the 50 students. Find the probability that this student wears trainers but does not wear glasses.
(iv) Two students are picked at random from those wearing trainers. Find the probability that both students have mobile phones.
▶️ Answer/Explanation
(a)(i) Ans: The tree diagram branches should be labeled with \( \frac{1}{6} \) for “Late” and \( \frac{5}{6} \) for “Not Late” on both Monday and Tuesday.
(a)(ii) Ans: \( \frac{5}{36} \)
Multiply the probability of being late on Monday (\( \frac{1}{6} \)) by the probability of not being late on Tuesday (\( \frac{5}{6} \)): \( \frac{1}{6} \times \frac{5}{6} = \frac{5}{36} \).
(b)(i) Ans: \( (G \cup T \cup M)’ \)
This represents the region outside all three circles (glasses, trainers, mobile phones).
(b)(ii) Ans: 28
Calculate \( T’ \cap (G \cup M) \): Total students (50) minus those wearing trainers (17) minus the one student outside all circles = 28.
(b)(iii) Ans: \( \frac{17}{50} \)
There are 17 students wearing trainers but not glasses. Probability = \( \frac{17}{50} \).
(b)(iv) Ans: \( \frac{4}{7} \)
From the 14 students wearing trainers with mobile phones, the probability of picking two is \( \frac{14}{17} \times \frac{13}{16} \), simplified to \( \frac{4}{7} \).
The diagram shows 5 cards.
(a) Donald chooses a card at random.
(i) Write down the probability that the number of dots on this card is an even number.
(ii) Write down the probability that the number of dots on this card is a prime number.
(b) Donald chooses two of the five cards at random, without replacement.
He works out the total number of dots on these two cards.
(i) Find the probability that the total number of dots is 5.
(ii) Find the probability that the total number of dots is an odd number.
▶️ Answer/Explanation
(a)(i) Ans: \(\frac{4}{5}\)
There are 4 even-numbered cards (2, 4, 6, 8) out of 5 total cards.
(a)(ii) Ans: \(\frac{4}{5}\)
There are 4 prime-numbered cards (2, 3, 5, 7) out of 5 total cards.
(b)(i) Ans: \(\frac{6}{20}\)
Total combinations: \( \binom{5}{2} = 10 \). Favorable pairs: (2,3), (3,2), (1,4), (4,1), (2,3), (3,2). Probability = \( \frac{6}{20} \).
(b)(ii) Ans: \(\frac{8}{20}\)
Odd totals occur when one card is odd and the other is even. There are 8 such combinations out of 20 possible.