Standard +0.3 This question requires recognizing that the mean of a binomial distribution (np = 4.8 with n = 20) gives p = 0.24, then calculating P(X < 3) = P(X ≤ 2) using binomial probability formula or tables. It's slightly easier than average as it's a straightforward two-step problem: derive the parameter, then apply standard binomial calculation with no conceptual complications.
1 A biased die was thrown 20 times and the number of 5 s was noted. This experiment was repeated many times and the average number of 5 s was found to be 4.8 . Find the probability that in the next 20 throws the number of 5 s will be less than three.
Summing 2 or 3 binomial probs o.e., any \(p\), \(n = 5\) or \(20\); Correct unsimplified answer; Correct answer
SR max 3 out of 4
B1, M1, A1
As above; Using \(N(4.8, 3.648)\) with cc \(2.5\) or \(3.5\) or \(0.114\) seen
$20p = 4.8$, $p = 0.24$ or $\frac{4.8}{20}$ | B1 | Correct value for $p$
$\text{P}(0, 1, 2) = (0.76)^{20} + {}^{20}C_1(0.24)(0.76)^{19} + {}^{20}C_2(0.25)^2(0.76)^{18} = 0.109$ | M1, A1, A1 [4] | Summing 2 or 3 binomial probs o.e., any $p$, $n = 5$ or $20$; Correct unsimplified answer; Correct answer
SR max 3 out of 4 | B1, M1, A1 | As above; Using $N(4.8, 3.648)$ with cc $2.5$ or $3.5$ or $0.114$ seen
1 A biased die was thrown 20 times and the number of 5 s was noted. This experiment was repeated many times and the average number of 5 s was found to be 4.8 . Find the probability that in the next 20 throws the number of 5 s will be less than three.
\hfill \mbox{\textit{CAIE S1 2011 Q1 [4]}}