11 Duncan has a fair 4-sided spinner. The spinner can land on 2, 3, 5 or 7
Duncan spins the spinner once.
(a)
(i) On the probability scale, mark with a cross (×) the probability that the spinner lands on 4
(ii) On the probability scale, mark with a cross (×) the probability that the spinner lands on an odd number.
(2)
Duncan spins the spinner again.
(b) Complete the table to show all the possible outcomes for the two spins of the spinner. Six outcomes have been done for you. (3, 2) means 3 on the first spin and 2 on the second spin.
Number on the second spin
2
3
5
7
Number on the first spin
2
(2, 2)
(2, 5)
3
(3, 2)
(3, 7)
5
7
(7, 3)
(7, 7)
(2)
(c) Find the probability that the number on the first spin is greater than the number on the second spin. (2)
Mark scheme (a)
Scheme
Marks
(i) × at 0
B1
(ii) × at \(\dfrac{3}{4}\)
B1
(2)
Mark scheme (b)
Scheme
Marks
M1
Correct table
A1
(2)
Notes
M1: For 3 correct entries
Mark scheme (c)
Scheme
Marks
M1ft
\(\dfrac{6}{16}\) oe
A1ft
(2)
Notes
M1ft: For \(\dfrac{6}{a}\) where \(6 \leqslant a \leqslant 16\) or
For \(\dfrac{b}{16}\) where \(1 \leqslant b \leqslant 16\) where \(a\) and \(b\) are integers
