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# counting & probability

1. Peter and 8 of his friends are divided into 3 groups for preparing a Christmas party. 1 of them decorates a Christmas tree, 4 of them prepare the Christmas dinner and the remaining 4 go to buy the presents. If Peter does not decorate the Christmas tree and 2 particular friends must be in the same group, find the number of ways of forming the groups.

2. A test consists of 6 multiple choice questions and each question has 4 options. Students have to answer all the questions. 1 mark will be awarded for each correct answer and no marks will be deducted for incorrect answers. The passing mark of the test is 4. Suppose a student answers all the questions by guessing randomly if he does not prepare for the test. 40 students take the test and 10 of them do not prepare for the test. Assume the student who have prepared for the test can pass the test. If 2 students are selected from the class at random, find the probability that both of them pass the test. Provided that: the probability that a student who forgets to prepare for the test passes the test is 0.0376.

Answer for Q1 is 120.

### 1 Answer

- ycLv 59 years agoFavorite Answer
1.

n(2 particular friends are in the same group)

=n(2 particular friends prepare dinner)+n(2 particular friends buy presents)

=2C2x7C2x5C4x1C1+2C2x7C2x5C4x1C1

=210

n(Peter decorates the tree and 2 friends are in the same group)

=1C1x(2C2x6C2x4C4+2C2x6C2x4C4)

=30

Required number of ways=210-30=180

2.

P(both pass the test)

P(both pass with preparation)+P(1 passes with preparation and 1 passes0.5722 by guess)

+P(both pass by guess)

=(30/40)(29/39)+2[(10/40)x0.0376](30/39)+[(10/40)x0.0376][(9/39)x0.0376]

=0.5722 (4 dp)

2012-07-23 22:13:35 補充：

Alt. solution to Q1

n(Peter and 2 friends : dinner)

=3C3x6C1x5C4x1C1=30

n(Peter and 2 friends: presents)

=3C3x6C1x5C4x1C1=30

n(Peter: dinner+2 friends:presents)

=1C1x6C3x2C2X3C2x1C1=60

n(Peter:presents+2 friends:dinner)

=1C1x6C3x2C2X3C2x1C1=60

req'd no. of ways = 30+30+60+60=180