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A small aeroplane has 14 seats for passengers. The seats are arranged in 4 rows of 3 seats and a back row of 2 seats (see diagram). 12 passengers board the aeroplane.

a) How many possible seating arrangements are there for the 12 passengers? Give your answer correct to 3 significant figures.

These 12 passengers consist of 2 married couples (Mr and Mrs Lin and Mr and Mrs Brown), 5 students and 3 business people.

b) The 3 business people sit in the front row. The 5 students each sit at a window seat. Mr and Mrs Lin sit in the same row on the same side of the aisle. Mr and Mrs Brown sit in another row on the same side of the aisle. How many possible seating arrangements are there?

c) If, instead, the 12 passengers are seated randomly, find the probability that Mrs Lin sits directly behind a student and Mrs Brown sits in the front row.

پاسخ تشریحی :
نمایش پاسخ

a) ${}^{14}{P_{12}}$

$ = 4.36 \times {10^{10}}$

b) business people $3! = 6$

students $5! = 120$

married couples ${}^3{P_2} \times 2 \times 2 = 24$

total ways $ = 17280$

c) Mrs Brown 3

Mrs Lin 10

Student 5

Prob $ = 3 \times {\text{1}}0 \times 5 \times {}^{11}{P_9}$/ (a)

$ = 0.0687$

OR1 $3/14 \times 10/13 \times 5/12 = 150/2184{\text{ }}\left( {0.0687} \right)$

OR2 $1 - 3/14 = 11/14$

$1 - 11/14 \times 5/13 = 127/182$

$8/14\left( {4/13 \times 12/12 + 9/13 \times 7/12} \right) + $

$3/14\left( {3/13 \times 12/12 + 10/13 \times 7/12} \right)$

$ = 1206/2184$

$1 - \left( {1524 + 1716 - 1206} \right)/2184 = 150/2184$

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