Question

there are 8 acts in a talent show.
an acrobat, a comedian, a dancer, a guitarist, a juggler, a singer, a violinist, and a whistler.
a talent show host randomly schedules the 8 acts.
compute the probability of each of the following events.
event a: the dancer is first, the comedian is second, the guitarist is third, and the whistler is fourth.
event b: the first four acts are the dancer, the juggler, the violinist, and the singer, in any order.
write your answers as fractions in simplest form.
$p(a) = \square$
$p(b) = \square$

Explanation:

Step1: Calculate total permutations

The total number of ways to arrange 8 acts is \(8! = 8\times7\times6\times5\times4\times3\times2\times1 = 40320\).

Step2: Probability of Event A

Event A: Dancer first, Comedian second, Guitarist third, Whistler fourth. The remaining 4 acts can be arranged in \(4!\) ways, but the first four positions are fixed. So the number of favorable arrangements for A is \(1\times1\times1\times1\times4! = 24\)? Wait, no—wait, Event A has specific first, second, third, fourth: dancer (1st), comedian (2nd), guitarist (3rd), whistler (4th). Then the remaining 4 acts (acrobat, juggler, singer, violinist) can be arranged in the last 4 positions: \(4! = 24\). Wait, no—wait, actually, for Event A, the first four positions are fixed (dancer 1st, comedian 2nd, guitarist 3rd, whistler 4th), so the number of favorable outcomes is \(1\times1\times1\times1\times4! = 24\)? Wait, no, the total number of permutations is \(8!\). Wait, no—actually, for Event A, the first position must be dancer (1 choice), second comedian (1 choice), third guitarist (1 choice), fourth whistler (1 choice), and the remaining 4 positions (5th to 8th) can be any of the remaining 4 acts: acrobat, juggler, singer, violinist. So the number of favorable arrangements is \(1\times1\times1\times1\times4! = 24\). Wait, but the total number of arrangements is \(8! = 40320\). So \(P(A) = \frac{4!}{8!}\)? Wait, no—wait, \(8! = 8\times7\times6\times5\times4!\), so \(\frac{4!}{8!} = \frac{1}{8\times7\times6\times5} = \frac{1}{1680}\)? Wait, no, wait: the first four positions are fixed (dancer, comedian, guitarist, whistler), so the number of favorable is 1 (for 1st) * 1 (2nd) * 1 (3rd) * 1 (4th) * 4! (for the last four). Wait, no, the last four positions: after fixing the first four, there are 4 acts left, so 4! ways. But the total number of arrangements is 8! So \(P(A) = \frac{4!}{8!}\)? Wait, no, 8! is 40320, 4! is 24, so 24 / 40320 = 1 / 1680? Wait, no, 8! = 8×7×6×5×4! = 1680×24, so 24 / (1680×24) = 1/1680? Wait, no, that's not right. Wait, actually, for Event A, the first position: probability dancer is first: 1/8. Then comedian is second: given dancer is first, 1/7. Then guitarist third: 1/6. Then whistler fourth: 1/5. Then the rest can be anything. So the probability is (1/8)×(1/7)×(1/6)×(1/5) = 1/(8×7×6×5) = 1/1680. Because after choosing dancer first (1/8), then comedian second (1/7 remaining), then guitarist third (1/6 remaining), then whistler fourth (1/5 remaining). So that's the probability.

Step3: Probability of Event B

Event B: The first four acts are dancer, juggler, violinist, singer in any order. So we need to choose 4 acts (dancer, juggler, violinist, singer) for the first four positions, and arrange them, then the remaining 4 acts (acrobat, comedian, guitarist, whistler) for the last four positions. The number of ways to arrange the first four acts: \(4!\) (since they can be in any order). The number of ways to arrange the last four acts: \(4!\). The total number of arrangements: \(8!\). So the number of favorable outcomes is \(4! \times 4!\). Therefore, \(P(B) = \frac{4! \times 4!}{8!}\). Let's compute that: \(4! = 24\), so numerator is 24×24=576. Denominator is 40320. Simplify: 576 / 40320. Divide numerator and denominator by 576: 576 ÷ 576 = 1, 40320 ÷ 576 = 70. Wait, 576×70 = 40320? 576×70 = 40320. So 576/40320 = 1/70? Wait, no: 8! = 40320, 4!×4! = 24×24=576. 576 / 40320 = (576 ÷ 576) / (40320 ÷ 576) = 1 / 70. Alternatively, using probability: the number of ways to choose the first four acts (dancer, juggler, violinist, singer) and arrange them: \(P(4,4) = 4!\) (permutations of 4 acts in 4 positions), then the remaining 4 acts: \(4!\). So total favorable: \(4! \times 4!\). Total permutations: \(8!\). So \(P(B) = \frac{4! \times 4!}{8!} = \frac{24 \times 24}{40320} = \frac{576}{40320} = \frac{1}{70}\) (dividing numerator and denominator by 576: 576 ÷ 576 = 1, 40320 ÷ 576 = 70). Alternatively, think of it as: first, choose 4 positions out of 8 for the first four acts? No, Event B says "the first four acts are dancer, juggler, violinist, singer in any order"—so the first four positions must be these four, in any order, and the last four positions are the remaining four. So the number of ways to arrange the first four: 4! (permutations of the four), and the last four: 4! (permutations of the remaining four). So total favorable: 4!×4!. Total permutations: 8!. So \(P(B) = \frac{4! \times 4!}{8!} = \frac{24 \times 24}{40320} = \frac{576}{40320} = \frac{1}{70}\) (since 40320 ÷ 576 = 70: 576×70 = 40320).

Wait, let's recheck Event A:

Event A: Dancer first (1/8), Comedian second (1/7, since one act is used), Guitarist third (1/6), Whistler fourth (1/5). Then the rest can be anything. So the probability is (1/8) × (1/7) × (1/6) × (1/5) = 1/(8×7×6×5) = 1/1680.

Event B: First four acts are dancer, juggler, violinist, singer in any order. So the number of ways to arrange these four in the first four positions: 4! (since 4 distinct acts, permutations). Then the remaining four acts (acrobat, comedian, guitarist, whistler) can be arranged in the last four positions: 4! ways. So total favorable: 4! × 4! = 24 × 24 = 576. Total permutations: 8! = 40320. So probability is 576 / 40320 = (576 ÷ 576) / (40320 ÷ 576) = 1 / 70.

Yes, that makes sense.

Answer:

\( P(A) = \boxed{\dfrac{1}{1680}} \)
\( P(B) = \boxed{\dfrac{1}{70}} \)