问题1.袋子A包含5个白色和6个黑色的球。另一个袋子B包含4个白球和3个黑球。将球从袋子A转移到袋子B,然后从袋子B中抽出一个球。找出被抽出的球是黑色的概率吗?
解决方案:
Bag ‘A’ contains 5 white balls and 6 black balls
Bag ‘B’ contains 4 white balls and 3 black balls
There are two ways of transferring ball:
1. Transfer 1 white ball from A to B and then draw a black ball from bag B
2. Transfer 1 black ball from A to B and then draw a black ball from bag B
Let E1, E2, A be events as follows:
E1: White ball drawn from bag 1
E2: Black ball drawn from bag 2
A: PROBABILITY OF BALL BEING BLACK
Step 1: Remove probability of drawing 1 white ball from bag A
EVENT 1(E1): One white ball drawn from bag A
P(E1) = 5/11 -(1)
Step 2: Remove probability of drawing 1 black ball from bag A
EVENT 2(E2): One black ball drawn from bag A
P(E2) = 6/11 -(2)
Step 3: Remove the probability that a ball drawn from bag B is black
with 1 white ball increased and 1 black ball increased
(A): Drawing one black ball from bag B
P(B/E1) = 3/8 -(E1 increased 1 white ball in bag B)(3)
P(B/E2) = 4/8 -(E2 has increased 1 black ball in bag B)(4)
Step 4: Apply the formula and put the removed values of events
By the law of probability
P(A) = P(E1)*P(A/E1)+P(E2)*P(A/E2)
= (5/11)*(3/8)+(6/11)*(4/8) -(From 1, 2, 3, 4)
= 39/88
The probability of ball being black is 39/88
问题2:一个钱包包含2个银币和4个铜币。另一个钱包包含4个银币和3个铜币。如果从2个钱包中随机抽取一个硬币,发现该硬币是银的可能性是多少?
解决方案:
Purse ‘1’ contains 2 silver coins and 4 copper coins
Purse ‘2’ contains 4 silver coins and 3 copper coins
One coin is from one of the purses and the coin is silver
Step 1: Finding probability of selecting purse 1
E1: Selecting purse 1
P(E1) = 1/2
Step 2: Finding probability of selecting purse 2
E2: Selecting purse 2
P(E2) = 1/2
A: Probability of drawing silver coin
P(A|E1) = 2/6
= 1/3 -(drawing silver coin from purse 1)
P(A|E2) = 4/7 -(drawing coin from purse 2)
From the law of probability
P(A) = P(E1)*P(A|E1)+P(E2)*P(A|E2)
= (1/2)*(1/3)+(1/2)*(4/7)
= (1/6)+(4/14)
= 19/42
The probability of drawing silver coin from two purses is 19/42
问题3:一个袋子包含4个黄色和5个红色球,另一个袋子包含6个黄色和3个红色球。球从袋子1转移到袋子2。找到从袋子2抽出的球是黄色的概率是多少?
解决方案:
Bag 1 contains 4 yellow and 5 red balls
Bag 2 contains 6 yellow and 3 red balls
A ball is transferred from bag1 to bag 2 this can be done in 2 ways:
1. Transferring yellow ball to bag 2
2. Transferring red ball to bag 2
Let E1, E2, and A be event such as:
E1: One yellow ball drawn from bag 1
E2: One red ball drawn from bag 2
A: One yellow drawn from bag 2
Step 1: Find the probability of event E1
P(E1) = 4/9 -(1)
Step 2: Find the probability of event E2
P(E2) = 5/9 -(2)
Step 3: Find the probability of drawing yellow ball from
bag 2 after transfer of balls from bag 1
P(A|E1) = 7/10 -(3) (Since E1 has increased 1 yellow ball in bag 2)
P(A|E2) = 6/10 -(4) (Since E2 has increased 1 red ball in bag 2)
Step 4: Applying the law of probability
P(A) = P(E1)*P(A|E1)+P(E2)*P(A|E2)
= (4/9)*(7/10)+(5/9)*(6/10) -(From 1, 2, 3, 4)
= 58/9
= 29/45
The probability of drawing yellow ball is 29/45
问题4:一个袋子包含3个白球和2个黑球。另一个袋子2包含2个白球和4个黑球。随机选择一个袋子,并从该袋子中抽出一个球。找出球变白的几率?
解决方案:
Bag 1 contains 3 white balls and 2 black balls
Bag 2 contains 2 white balls and 4 black balls
One bag is chosen at random
Let E1, E2, and A be events such as
E1: Selecting bag I
E2: Selecting bag II
A: Drawing white ball
Step 1: Probability of selecting bag 1
P(E1) = 1/2 -(1)
Step 2: Probability of selecting bag 2
P(E2) = 1/2 -(2)
Step 3: Find the probability of drawing white ball
P(A|E1) = P (Drawing white ball from bag 1)
= 3/5 -(iii)
P(A|E2) = P (Drawing white ball from bag 2)
= 2/6 = 1/3 (iv)
Step 4: Use the law of probability
P(A) = P(E1)*P(A|E1)+P(E2)*P(A|E2) -(From 1, 2, 3, 4)
= (1/2)*(3/5)+(1/2)*(1/3)
= 7/15
The probability of ball being white is 7/15
问题5:3袋的内容如下
包I:1个白色,2个黑色,3个红色球
第二袋:2个白色,1个黑色,1个红球
第三类包:4个白色,5个黑色,3个红球
随机选择一个袋子并抽出两个球,那么球变成红色和白色的概率是多少?
解决方案:
Let E1, E2, E3, A be the events such as
E1: Selecting BAG I
E2: Selecting BAG II
E3: Selecting BAG III
A: Drawing a red and a white ball
Step 1: Finding the probability of selecting one bag at random
P(E1): 1/3 -(1)
P(E2): 1/3 -(2)
P(E3): 1/3 -(3)
Step 2: Finding the probability of drawing one red and one white ball
P(A|E1) = P (Drawing one red and one white ball from bag I)
= 1C1 * 3C1/6C2 (nCr = n!/r!*(n-r)!)
= (2 * 1) / (4 * 3) /2
= 1/5 -(4)
P(A|E2) = P(Drawing ball 1 red and white ball from bag II)
= (2*1)/4*3/2
= 1/3 -(5)
P(A|E3) = P(Drawing ball 1 red and white ball from bag III)
= 2/11 -(6)
Step 3: Applying the law of probability
P(A)= P(E1)*P(A|E1)+P(E2)*P(A|E2)+P(E3)*P(A|E3) -(From 1, 2, 3, 4, 5, 6)
= (1/3*1/5)+(1/3*1/3)+(1/3*2/11)
= (1/15)+(1/9)+(2/33)
= 118/495
The required probability is 118/495
问题6:掷无偏硬币是结果,结果是将一对无偏骰子掷出,并记下了获得的数字总和。如果结果为12,则从11张纸牌中取出一张纸牌,编号为2,3,4,5…..12,并记下许多纸牌。这个数字是7还是8的概率是多少?
解决方案:
An unbiased coin is tossed then:
1. If the head occurs a pair of dice is rolled and the sum of them is either 7 or 8
2. If tail occurs a card is drawn from card numbered 2,3,4,5 ….. 12. And is 7 or 8
Let E1, E2, & A be events as
E1: A head occurs
E2: A tail occurs
A: The noted number is 7 or 8
Step 1: Find the probability of tail comes up in a single coin
P(E1) = 1/2 -(1)
Step 2: Find the probability of head comes up in a single coin
P(E2) = 1/2 -(2)
Step 3: Probability that number noted is 7 or 8
P(A|E1) = P[Pair of dice shows 7 or 8 as sum]
(Sum of pair is 7 or 8 when – (1,6), (2,5), (3,4),
(4,3), (5,2), (6,1), (6,2), (5,3), (4,4), (3,5), (2,6))
P(A|E1) =11/36 -(3)
P(A|E2) = P[7 or 8 occurs on cards drawn numbered 2,3,4….12]
= 2/11 -(4)
Step 4: Applying the law of probability
P(A) = P(E1)*P(A|E1)+P(E2)*P(A|E2)
= (1/2)*(11/36)+(1/2)*(2/11)
=193/792
The required probability is 193/792
问题7.一个工厂有两台机器A和B。过去的记录显示,机器A生产了60%的产出项目,而机器B生产了40%的产出项目。此外,机器A生产的2%的物品有缺陷,而机器B生产的1%的物品有缺陷。如果随机抽取物品,则该物品有缺陷的可能性是多少?
解决方案:
Let E1, E2, A be the events such as
E1: Item produced by machine A
E2: Item produced by machine B
A: Product is defective
Step 1: Remove the probability of E1 and E2
P(E1) = 60%
= 60/100 -(1)
P(E2) = 40%
= 40/100 -(2)
Step 2: Find the probability that the product is defective
P(A|E1) = P(Defective item from machine A)
= 2%
= 2/100 -(3)
P(A|E2) = P(Defective item from machine B)
= 1%
1/100 -(4)
Step 3: Applying the law of probability
P(A) = P(E1)*P(A|E1)+P(E2)*P(A|E2)
= (60/100)*(2/100)+(40/100)+(1/100) (From 1, 2, 3, 4)
= 160/10000
= 0.016
The required probability is 0.016
问题8.袋子A包含8个白色和7个黑色的球,袋子B包含5个白色和4个黑色的球。从袋子A中捡起一个球,并与袋子B中的球混合。从袋子B中随机抽出一个球。该球是白色的概率是多少?
解决方案:
Transfer if the ball from bag A to bag B can be done in two ways:
1. One white ball is transferred from bag A to B, then a white ball is drawn
2. One black ball is transferred from bag A to B, then a black ball is drawn.
Let E1, E2 & A be the events as:
E1: One white ball from bag A
E2: One black ball from bag A
A: One white ball from bag B
Step 1: Find the probability of balls drawn from bag A
P(E1) = 8/15 -(1)
P(E2) = 7/15 -(2)
Step 2: Find the probability of a ball drawn from bag B
P(A|E1) = 6/10 -(3) (Since E1 has increase white ball by 1 in bag B)
P(A|E2) = 4/10 -(4) (E2 increase black ball in bag 2)
Step 3: Applying the law of probability
P(A) = P(E1)*P(A|E1)+P(E2)*P(A|E2)
= (8/15)*(6/10)+(7/15)*(4/10) -(From 1, 2, 3, 4)
= 83/150
The required probability is 83/150
问题9.袋子1包含4个白色和5个黑色的球,另一个袋子包含3个白色和4个黑色的球。取出一个没有看到颜色的球,将其放在另一个袋子中。稍后将球从中取出。找出球是白色的几率?
解决方案:
A ball is taken out from bag I and put in bag II without seeing its color.
Then the ball is taken out from bag II and it is white
Step 1: Find the probability that the ball is drawn from bag 1 is white
P(W1) = 4/9 -(1)
Step 2: Find the probability that the ball is drawn from bag 1 is black
P(B1) = 5/9 -(2)
Step 3: Find the probability after all is moved to bag 2
P(W2/B1) = P (White ball drawn from bag 2 after B1 transfer)
= (3/8) -(3)
P(W2/W1) = P(White ball drawn from bag 2 after W1 transfer)
= (4/8)
= (1/2) -(4)
Step 4: Applying the law of probability
P(White ball from bag 2)= P(B1)*P(W2/B1)+P(W1)*P(W2/W1)
= (5/9)*(3/8)+(4/9)*(1/2)
= 31/72
The required probability 31/72
问题10.一个袋子包含4个白色和5个黑色的球,另一个袋子包含6个白色和7个黑色的球。将球从袋子I转移到袋子II,并将球从袋子II抽出。找出球是白色的几率?
解决方案:
A ball is taken from bag 1 and without seeing the color the ball is mixed with balls in bag 2. Then the ball is drawn from the bag 2, what is the probability that the ball is white
Step 1: Find the probability that the ball removed from bag 1 is white
P(1 White ball from bag 1) = P(W1)
= 4/9 -(1)
Step 2: Find the probability that the ball removed from bag 1 is black
P(1 Black ball drawn from bag 2) = P(B1)
= 5/9 -(2)
Step 3: Find the probability that the ball drawn from bag 2 is white after transfer
P(White ball transferred from bag 1 to bag 2) = P(W2/W1)
= 7/14
= 1/2 -(3)
P(Black ball transferred from bag 1 to bag 2) = P(W2/B1)
= 6/14
= 3/7 -(4)
Step 4: Applying the law of probability
P(1 WHITE FROM BAG II) = P(W1)*P(W2/W1)+P(B1)*P(W2/B1)
= (4/9)*(1/2)+(5/9)*(3/7) (From 1, 2, 3, 4)
= 58/126
= 29/63
The required probability 29/63
问题11.包含10个白球和3个黑球。另一个包含3个白球和5个黑球。从骨灰盒I中抽出两个球,放入第二个骨灰盒中,然后再抽出一个球。找出球是白色的几率?
解决方案:
URN “1”
10 white balls (10 WB) 3 black balls (3 BB) |
URN “2”
3 white balls (3 WB) 5 black balls (5 BB) |
Let U12W, U11W1B, U1 2B be the events of transferring 2 white balls (2W), one white & black ball(1W1B), and transferring 2 black balls 2BB from urn I to urn II.
P(drawing two white balls from urn I) = 10C2 /13C2 = 45/78 -(1)
P(drawing 1 white and 1 black ball from urn I) = 10C1 3C1/13C1 = 30/78 -(2)
P(drawing 2 black balls from urn I) = 3C2/13C2 = 3/78 -(3)
Let U2(w) drawing from the white ball from urn II. There will be three cases based on above events
I | II | III | |
5 WHITE | 4 WHITE | 3 WHITE | |
5 BLACK | 6 BLACK | 7 BLACK | |
TOTAL | 10 | 10 | 10 |
P(Based on event 1BASED ON EVENT 1) = 5C1/10C1
= 5/10
= 1/2 -(4)
P(Based on event 2) = 4C1/10C1
= 4/10
= 2/5 -(5)
P(Based on event 3) = 3C1/10C1
= 3/10 -(6)
From the law of probability
P(U2W) = (45/78)*(1/2)+(30/78)*(2/5)+(3/78)*(3/10) -(From 1, 2, 3, 4, 5, 6)
= 59/130
The required probability 59/130
问题12.一个袋子包含6个红色和8个黑色的小球,另一个袋子包含8个红色和6个黑色的小球。从袋子I中抽出一个球,而袋子II中看不到颜色。然后从袋子II中抽出球,找到球变红的概率?
解决方案:
Given: Bag I contain 6 red(R1) and 8 black(B1) balls
Bag II contains 8 red(r2) and 6 black(B2) balls
A ball is drawn from bag I and without seeing color is kept in bag II,
then the ball drawn from bag 2 is Red
P(One red ball from bag 2)
= P((B1 n R2)U (R1 n R2))
= P(B1 n R2) + P(R1 n R2)
= P(B1)*P(B1/R2)+P(R1)*P(R1/R2)
= (8/14)*(8/15)+(6/14)*(9/15)
= 118/210
= 59/105
The required probability is 59/105
问题13:三台机器E1, E2, E3分别产生电灯泡日总产量的50%,25%和25%。众所周知,每台机器E1和E2所生产的每根管子中有4%是有缺陷的,而在E3上生产的那些管子中有5%是有缺陷的。如果从一天的生产中挑出一根管子,那么计算出该管子有缺陷的概率是多少?
解决方案:
Let D be the event that the bulb is defective
A1, A2 & A3 be the events that tube is produced from the machine E1, E2, & E3
P(D) = P(A1)*P(D|A1)+P(A2)*P(D|A2) -(1)
P(A1) = 50/100
= 1/2 -(2)
P(A2) = 25/100
= 1/4 -(3)
P(A3) = 25/100
= 1/4 -(4)
P(D|A1) = 4/100 = P(D|A2)
= 1/25 -(5)
P(D|A3) = 5/100
= 1/20 -(6)
By applying the law of probability:
P(D) = (1/2)*(1/25) + (1/4)*(1/25) + (1/4)*(1/25)
= 17/400