SL 4.5—Probability concepts, expected numbers

Numbers greater than 4 on a six-sided die are $\{5,6\}$, so there are 2 favourable outcomes out of 6 equally likely outcomes.

$P(\text{greater than 4})=\frac{2}{6}=\frac{1}{3}$

Using $E=P(A)\times n$: $E=\frac{1}{3}\times 180=60$

The expected number of rolls greater than 4 is 60.

We need $E=P(A)\times n$ where $n=180$ and $A$ is rolling greater than 4.

30 would correspond to $P=\frac{1}{6}$, which is the probability of rolling exactly one specific face — but we need two faces $\{5,6\}$, so this is too small.

45 corresponds to $P=\frac{45}{180}=\frac{1}{4}$, which does not match any natural grouping of faces on a fair die.

90 corresponds to $P=\frac{1}{2}$, which would mean 3 favourable outcomes — but only $\{5,6\}$ are greater than 4, giving 2 outcomes, not 3.

Only 60 correctly reflects $P=\frac{2}{6}=\frac{1}{3}$ applied to 180 trials.

The probability of a defective item is $P(\text{defective})=0.08$ and the number of trials is $n=1250$.

$E=P(A)\times n=0.08\times 1250$

$E=0.08\times 1250=100$

The expected number of defective items is 100.

We compute $E=0.08\times 1250$ and compare with each option.

80 would require $P=\frac{80}{1250}=0.064$, which does not match the given $P=0.08$.

108 corresponds to $P=\frac{108}{1250}=0.0864$, a common rounding error — not equal to $0.08$.

156 corresponds to $P\approx 0.125$, which could arise from incorrectly using $\frac{1}{8}$ instead of $0.08$.

100 is the only value consistent with $0.08\times 1250=100$.