Definition of probability

Exercise 2

1. When throwing one die 6 times, find the following probability．

(a)

one pit appears once．

(b)

one pit appears 4 times．

(c)

one pit appears less than or equal to 4 times．

(d)

one pit appears more that or equal to 5 times.．

2. One bag contains 5 white balls, 3 red balls, and 2 black balls. When taking out four balls from them, find the following probabilities.

(a)

everything is white．

(b)

there are exactly two whites．

(c)

No more than 2 whites．

(d)

2 whites, 2 reds．

(e)

white, red, and black are all included

3. There are cards numbered from 1 to 10. When arranging these cards in a row, find the following probabilities

(a)

When 1 to 10 are arranged in that order

(b)

the 4 card is just 4th

(c)

1 is first and 4 is fourth

4. Suppose that a lot of 8 cm squares are drawn on the graph paper at intervals of 8 cm vertically and horizontally. When throwing a disk with a diameter of 3 cm, find the following probability．

(a)

The disk fits inside one square．

(b)

Disk touches the side of the square．

(c)

Disk touches four squares．

5. There is a bag containing 4 white balls and 6 red balls. Find the following probabilities when taking out two pieces from this bag at the same time.．

(a)

the probability of two balls are white

(b)

probability that only one is a white ball

(c)

probability that at least one is a white ball

Answer

(a) Find the probability using the number of cases. $X_{i} =$[1 pit i times out of 6 times].

If you throw the dice 6 times, there are a total of ${}_6 \Pi_{6} = 6^6$. Next, let's count how many times there are cases where the 1st roll appears once out of 6 times．

If you get a 1 on the first throw, the remaining 5 times are other than 1. Then ${}_5 \Pi_{5}$．Similarly for 2nd, 3rd,4th, Thus,

$${}_6 C_{1} \cdot {}_5 \Pi_{5} = 6 \cdot 5^5$$

Therefore the probability that 1 roll out of 6 times is

$$P_{r}(X_{1}) = \frac{6 \cdot 5^5}{6^6} = \left(\frac{5}{6}\right)^5$$

(b) There are a total of ${} _ 6 C_ {4}$ combinations in which 1's are rolled 4 times. At that time, the remaining two times are other than 1, so $5 ^ 2$. So in total

$${}_6C_{4} \cdot 5^2$$$$\mbox {ways}$$

Therefore, the probability that a 1's roll will appear 4 out of 6 times is

$$P_{r}(X_{4}) = \frac{{}_6 C_{4} \cdot 5^2}{6^6} = \frac{375}{46656}$$

(d) The number of 1's that appear 5 times or more is

$$X_{5} \cup X_{6}$$

The probability that a 1 will appear 4 times or less is

$$P_{r}\left(X_{0} \cup X_{1} \cup X_{2} \cup X_{3} \cup X_{4}\right) = \frac{46625}{46656}$$

Then，

$$P_{r}(X_{5} \cup X_{6}) = 1 - P_{r}\left(X_{0} \cup X_{1} \cup X_{2} \cup X_{3} \cup X_{4}\right) = \frac{31}{46656}$$

2. (a) There is no choice but to take out 4 out of the 5 white balls in the bag，The combination is ${}_5 C_{4}$． Therefore, the probability that all four taken out are white balls is

$$\frac{{}_5 C_{4}}{{}_{10} C_{4}} = \frac{\frac{5!}{4! 1!}}{\frac{... ... = \frac{5 \cdot 4 \cdot 3 \cdot 2}{10 \cdot 9 \cdot 8 \cdot 7} = \frac{1}{42}$$

(b) Two balls are white when 2 white balls are selected from 5 white balls and two balls are from red and black balls. Thus the number of outcomes is ${}_5 C_{2} \times {}_5 C_{2}$．Therefore, the probability that white is exactly 2 out of the 4 taken out is

$$\frac{{}_5 C_{2} \times {}_5 C_{2}}{{}_{10} C_{4}} = \frac{10}{21}$$

2 (c) The case of 2 whites or less is the case of 0, 1 or 2 white events. Therefore, the probability is

$$\frac{{}_5 C_{0} \cdot {}_5 C_{4} + {}_5 C_{1} \cdot {}_5 C_{3} + {}_5 C_{2} \times {}_5 C_{2}}{{}_{10} C_{4}} = \frac{31}{42}$$

(d) The combination of two whites and two reds is ${}_5 C_{2} \times {}_3C_{2}$. Therefore, the probability that white is 2 and red is 2 out of the 4 taken out is

$$\frac{{}_5 C_{2} \cdot {}_3 C_{2}}{{}_{10} C_{4}} = \frac{1}{7}$$

(e) Pay attention to the 4 balls taken out．

1． The fact that white, red, and black are all included means that one of white, red, and black can be considered as a combination of two. Thus ${4 \choose 2,1,1}$．

2． The probability of taking out one white, one red, one black is $\frac{5}{10}\frac{3}{9}\frac{2}{8}$．

3．Therefore, the probability that white, red, and black are included is

$${4 \choose 2,1,1}\frac{5}{10}\frac{3}{9}\frac{2}{8} = \frac{1}{2}$$

3.

(a) Let's think about how many ways to arrange 1 to 10 in a row. There can be any of 1 to 10 at the beginning, so there are 10 ways，Next is 9 ways $\ldots$，The total is ${}_{10} P_{10} = 10!$ ways.

Thus the probability of 1 to 10 are lined up in that order is

$$\frac{1}{10!}$$

(b) The fact that the 4 card is exactly the 4th means that the other 9 cards can be anywhere, so the 4th card is exactly the 4th card in ${} _ 9 P_ {9} = 9!$ Street There is. Therefore, the probability is

$$\frac{{}_9 P_{9}}{{}_{10} P_{10}} = \frac{9!}{10!} = \frac{1}{10}$$

(c) The fact that 1 is first and 4 is fourth means that the other eight cards can be anywhere, so there are a total of ${}_ 8 P_{8}$ ways. Therefore, the probability that 1 will come first and 4 will come fourth is

$$\frac{{}_8 P_{8}}{{}_{10} P_{10}} = \frac{8!}{10!} = \frac{1}{10\cdot 9} = \frac{1}{90}$$

4. (a) The radius of the disk is more than 1.5 cm. To fit inside a square, the center of the disk should be inside a square with a side of 5 cm. Therefore, the probability is

$$\frac{\mbox{area of 5cm side square}}{\mbox{area of 8cm side square}} = \frac{25}{64}$$

(b) The complement of A = 「The disk touches the side of the square」is $\bar{A}$ = 「The disk is inside the square．Thus the probability is，

$$1 - \frac{\mbox{area of 5cm side square}}{\mbox{area of 8cm side square}} = \frac{39}{64}$$

(c) In order for a disk to span four squares, its center must be within 1.5 cm of the boundary of the four squares. The area is $\ pi (1.5) ^ 2$. Therefore, the probability that a disk will span four squares is

$$\frac{\pi (1.5)^2}{64}$$

5.

(a) There are only two out of the four white balls in the bag, so the combination is ${}_4 C_{2}$. Therefore, the probability that the two taken out are both white balls is

$$\frac{{}_4 C_{2}}{{}_{10} C_{2}} = \frac{\frac{4!}{2! 2!}}{\frac{10!}{8! 2!}} = \frac{4 \cdot 3}{10 \cdot 9} = \frac{2}{15}$$

(b) Only one white ball is the number when one is taken out from four white balls and one is taken out from six red balls.Then ${}_4 C_{1} \times {}_6 C_{1}$．Thus,

$$\frac{{}_4 C_{1} \times {}_6 C_{1}}{{}_{10} C_{2}} = \frac{4 \times 6}{45} = \frac{8}{15}$$

(c) The event that at least one is a white ball is either both white balls or only one white ball. Note that these events are mutual exclusivity events (which do not occur at the same time), in total.， ${}_4 C_{2} + {}_4 C_{1} \times {}_6 C_{1}$ ways．Thus the probability is

$$\frac{{}_4 C_{2} + {}_4 C_{1} \times {}_6 C_{1}}{{}_{10} C_{2}} = \frac{2}{15} + \frac{8}{15} = \frac{2}{3}$$