Topics › KS4 › Probability · Foundation & Higher
Combined events
-
A fair coin is flipped and a fair six-sided dice is rolled. What is the probability of getting a head and a 6?
A
Option A:
$\frac{2}{3}$
B
Option B:
$\frac{2}{8}$
C
Option C:
$\frac{1}{8}$
D
Option D:
$\frac{1}{12}$
Answer and misconceptions
A
Option A:
$\frac{2}{3}$
✗
Adds the probabilities for "and"Works out $\frac{1}{2} + \frac{1}{6}$; for independent events "and" means multiply: $\frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$.
B
Option B:
$\frac{2}{8}$
✗
Adds numerators and denominatorsWorks out $\frac{1 + 1}{2 + 6}$; fractions are not combined this way.
C
Option C:
$\frac{1}{8}$
✗
Adds the numbers of outcomesCounts $2 + 6 = 8$ outcomes; there are $2 \times 6 = 12$ equally likely pairs.
D
Option D:
$\frac{1}{12}$
✓ Correct
-
A spinner can land on red, blue or green. $P(\text{red}) = 0.3$ and $P(\text{blue}) = 0.25$. What is the probability it lands on red or blue?
A
Option A:
$0.075$
B
Option B:
$0.45$
C
Option C:
$0.55$
D
Option D:
$0.475$
Answer and misconceptions
A
Option A:
$0.075$
✗
Multiplies for "or"Works out $0.3 \times 0.25$; the spinner cannot land on both, so "or" means add.
B
Option B:
$0.45$
✗
Finds the probability of neitherWorks out $1 - 0.55$, which is the probability of green.
C
Option C:
$0.55$
✓ Correct
D
Option D:
$0.475$
✗
Subtracts P(both) when the events cannot both happenWorks out $0.3 + 0.25 - 0.3 \times 0.25$; the spinner cannot land on red and blue at once, so just add: $0.55$.
-
A bag has 5 red counters and 3 blue counters. Two counters are taken at random without replacement. What is the probability that both are red?
A
Option A:
$\frac{25}{64}$
B
Option B:
$\frac{5}{14}$
C
Option C:
$\frac{20}{64}$
D
Option D:
$\frac{25}{56}$
Answer and misconceptions
A
Option A:
$\frac{25}{64}$
✗
Treats the draws as with replacementWorks out $\frac{5}{8} \times \frac{5}{8}$; after one red is taken there are 4 red out of 7.
B
Option B:
$\frac{5}{14}$
✓ Correct
C
Option C:
$\frac{20}{64}$
✗
Reduces the red count but not the totalWorks out $\frac{5}{8} \times \frac{4}{8}$; there are only 7 counters left for the second draw.
D
Option D:
$\frac{25}{56}$
✗
Reduces the total but not the red countWorks out $\frac{5}{8} \times \frac{5}{7}$; after one red is taken only 4 reds are left: $\frac{5}{8} \times \frac{4}{7} = \frac{5}{14}$.
-
The probability that a bus is late on any day is $0.2$, independently of other days. What is the probability that it is late on at least one of two days?
A
Option A:
$0.36$
B
Option B:
$0.4$
C
Option C:
$0.04$
D
Option D:
$0.64$
Answer and misconceptions
A
Option A:
$0.36$
✓ Correct
B
Option B:
$0.4$
✗
Adds the probabilities for each dayWorks out $0.2 + 0.2$; this counts "late both days" twice. Use $1 - P(\text{not late both days}) = 1 - 0.8^2$.
C
Option C:
$0.04$
✗
Finds the probability of late on both daysWorks out $0.2 \times 0.2$; "at least one" also includes late on just one day.
D
Option D:
$0.64$
✗
Forgets to subtract from 1$0.8 \times 0.8 = 0.64$ is the probability it is never late; at least once is $1 - 0.64$.
-
The probability of rain on Saturday is $0.3$. The probability of rain on Sunday is $0.4$. The two events are independent. What is the probability that it rains on Saturday but not on Sunday?
A
Option A:
$0.12$
B
Option B:
$0.9$
C
Option C:
$0.18$
D
Option D:
$0.28$
Answer and misconceptions
A
Option A:
$0.12$
✗
Ignores the "not"Works out $0.3 \times 0.4$, rain on both days; no rain on Sunday has probability $0.6$.
B
Option B:
$0.9$
✗
Adds for "and"Works out $0.3 + 0.6$; for independent events, multiply: $0.3 \times 0.6 = 0.18$.
C
Option C:
$0.18$
✓ Correct
D
Option D:
$0.28$
✗
Applies "not" to the wrong dayWorks out $0.7 \times 0.4$, which is no rain on Saturday and rain on Sunday.