F.5 math on probability

呢啲加奶

2013-02-26 4:52 am

可不可以列式，如果可以就解釋一下。特別是Q1(b)，2，3，4，5。
1. In a bag, there are some red, green and orange jelly beans in the ratio 1:1:2.
If 2 jelly beans are chosen one by one at random with replacement, find the
probabilities that
(a) 2 orange jelly beans are chosen (ans: 1/4)
(b) at least one orange jelly bean id chosen (ans:3/4)

2. The dartboard in the figure is made up of 3 similar triangles, where
AC: DF: GI = 7: 5: 4. If Anton throws a dart at random and hits the dartboard,
find the probability that he hits the green region. (ans: 9/49)

3. A four- digit integer is formed by arranging the digit 3,4,5 and 6 randomly, If
each digit can be used once only, find the probabilities that the four-digit integer
(a) is a prime number. (ans: 0)

4. The music club of a secondary school gave away 100 tickets of their annual
performance to 100 students at random inthe first break. They gave away another
100 tickets to 100 students at random in the secomd break. If 8 of these 100
students had been given a ticket in the first break, estimate the total number of
students in the school. (ans:1250)

5. Five balls numbered from 1 to 5 are put into a box. John takes three balls at
random from the box first and Ken takes the remaning two balls. Find the
probability that the sum of the numbers of John's balls is greater than that of
Ken's. (ans: 4/5)

回答 (1)

不用客氣

2013-02-26 5:53 am

✔ 最佳答案

1.
(a)
O : A jelly bean is orange
N : A jelly bean is NOT orange

P(O) = 2/(1+1+2) = 1/2
P(N) = 1 - (1/2) = 1/2

The required probability
= P(O,O)
= P(O) x P(O)
= (1/2) x (1/2)
= 1/4

(b)
The required probability
= 1 - P(N,N)
= 1 - P(N) x P(N)
= 1 - (1/2) x (1/2)
= 3/4

Alternative method :
The required probability
= P(O,O) + P(O,N) + P(N,O)
= (1/4) + P(O) x P(N) + P(N) x P(O)
= (1/4) + (1/2) x (1/2) + (1/2) x (1/2)
= 3/4

=====
2.
What is the figure.
In the absence of the figure, no one can really answer the question.

A guessed answered is as follows :

Area ratio of the three similar triangles
= 7² : 5² : 4²
= 49 : 25 : 16

Let 49k, 25k and 16k be the areas of the three triangles.

Total area
= 49k

Area of the green region
= 25k - 16k
= 9k

The required probability
= 9k / 49k
= 9/49
(Just a guess !)

=====
3.
(a)
Sum of the 4 digits
= 3 + 4 + 5 + 6
= 18
= 3 x 6

Since the sum of the 4 digits is a multiple of 3, then the number formed byarranging the 4 digits must be the multiple of 3, and thus must not be a primenumber.
Hence, the required probability
= 0

=====
4.
Let n be the number of students in the school.

Proportion of the students who have been given a ticket in the first break :
8/100 ≈ 100/n
8n ≈ 10000
n ≈ 1250

The estimated number of students in the school = 1250

=====
5.
No. of ways that John takes three balls randomly and Ken takes the remainingtwo balls
= 5P3 x 2C2
= (5!/3!2!) x (2!/2!0!)
= 10 x 1
= 10

Sum of the 5 numbers
= 1 + 2 + 3 + 4 + 5
= 15

If the sum of numbers of John's balls is NOT greater than that of Ken's, thesum of numbers of Ken's ball must be at least 8.
Then, Ken's balls may be : (3, 5), (4, 5)
Number of ways = 2

P(sum of the numbers of John's balls is NOT greater than that of Ken's)
= 2/10
= 1/5

The required probability
= 1 - (1/5)
= 4/5

參考: 不用客氣

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