问题13:从包含5张纸牌的1、1、2、2和3的盒子中随机选择两张纸牌。让X代表总和,让Y代表抽取的两个数字的最大值。求出X和Y的概率分布,均值和方差。
解决方案:
As box contains cards numbered as 1,1,2,2 and 3
∴ possible sums of card numbers are 2,3,4 and 5
Hence, X can take values 2,3,4 and 5
X=2 [ when drawn cards are (1,1)]
X=3 [when drawn cards are (1,2) or (2,1)]
X=4 [when drawn cards are (2,2) or (3,1) or (1,3)]
X=5 [when drawn cards are (2,3) or (3,2)]
As Y is a random variable representing maximum of the two numbers drawn
∴ Y can take values 1,2 and 3.
Y=1 [when drawn cards are 1 and 1]
Y=2 [when drawn cards are (1,2) or (2,2) or (2,1)]
Y=3 [when drawn cards are (1,3) or (3,1) or (2,3) or (3,2)]
Note : P(1) represents probability of drawing card numbered as 1, similarly P(2) and P(3)
∴ P(X=2) = P(1)P(1) = 2/5 x 1/4 = 0.1
[For drawing first card we had 2 favourable outcomes as 1,1 out of total 5 ,in second time of drawing ,as we drew a card numbered as 1 we are having 1 favourable outcome out of total remaining of 4]
Similarly,
P(X=3) = P(2)P(1) + P(1)P(2) = 2/5 x 2/4 + 2/5 x 2/4 = 0.4
P(X=4) = P(2)P(2)+P(3)P(1)+P(1)P(3) = 2/5 x 1/4 + 2/5 x 1/4 + 1/5 x 2/4 = 0.3
P(X=5) = P(2)P(3)+P(3)P(2) = 2/5 x 1/4 + 2/5 x 1/4 = 0.2
Similarly,
P(Y=1) = P(1)P(1) = 2/5 x 1/4 = 0.1
P(Y=2) = P(1)P(2)+P(2)P(1)+P(2)P(2) = 2/5 x 2/4 + 2/5 x 2/4 + 2/5 x 1/4 = 0.5
P(Y=3) = P(2)P(3)+P(3)P(2)+ P(1)P(3)+P(3)P(1)
= 2/5 x 1/4 + 2/5 x 1/4 + 2/5 x 1/4 +2/5 x 1/5 + 1/5 x 2/4 = 0.4
Now we have pi and xi.
Let’s proceed to find mean and variance.
Mean of any probability distribution is given by Mean = ∑xipi
Variance is given by:
Variance = ∑ xi2pi – (∑xipi)2
∴ first we need to find the products i.e. pixi and pixi2 and add them to get mean and apply the above formula to get the variance.
Following table representing probability distribution gives the required products :
xi | pi | xipi | xi2pi |
0 | 1/16 | 0 | 0 |
1 | 7/16 | 7/16 | 7/16 |
2 | 5/16 | 10/16 | 20/16 |
3 | 2/16 | 6/16 | 18/16 |
∴ Mean for(X) = 0.2+1.2+1.2+1 = 3.6
Variance for(X) = 0.4+3.6+4.8+5.0-3.62 = 13.8-3.62 = 0.84
Similarly probability distribution for Y is given below:
yi | pi | yipi | yi2pi |
1 | 0.1 | 0.1 | 0.1 |
2 | 0.5 | 1.0 | 2.0 |
3 | 0.4 | 1.2 | 3.6 |
∴ Mean for(Y) = 0.1+1.0+1.2 = 2.3
Variance for(Y) = 0.1+2.0+3.6-2.32 = 5.7-2.32 = 0.41
问题14:一次掷骰子两次。 “成功”折腾得到一个奇数。查找成功次数的方差。
解决方案:
As success is considered when we get an odd number when we roll a die.
As die is rolled twice , so we can get no success or a single success or we can get odd both the times an odd number.
If X is the random variable denoting the success then X can take value 0,1 or 2
∵ P(getting an odd number in a single rolling of die) = 3/6 = 1/2
As rolling a die is an independent event:
∴ P(getting an odd on first roll and probability of getting odd on second roll)=P(getting an odd on first roll) x P(getting an odd on second roll)
Note: P(AՈB) = P(A)P(B) where A and B are independent events.
∴ P(X=0) = P(even number on first throw) x P(even on second throw) = 1/2 x 1/2 = 1/4
P(X=1) = P(even number on first throw) x P(odd on second throw) +
P(odd number on first throw) x P(even on second throw) = 1/2 x 1/2 + 1/2 x 1/2 = 1/2
P(X=2) = P(odd number on first throw) x P(odd on second throw) = 1/2 x 1/2 = 1/4
Now we have pi and xi.
Let’s proceed to find mean and variance.
Mean of any probability distribution is given by Mean = ∑xipi
Variance is given by:
Variance = ∑ xi2pi – (∑xipi)2
∴ first we need to find the products i.e. pixi and pixi2 and add them to get mean and apply the above formula to get the variance.
Following table representing probability distribution gives the required products :
xi | pi | xipi | xi2pi |
0 | 1/4 | 0 | 0 |
1 | 1/2 | 1/2 | 1/2 |
2 | 1/4 | 1/2 | 1 |
∵ Variance = ∑ xi2pi – (∑xipi)2
∴ Variance = 0 + 1/2 + 1 – (0 + 1/2 + 1/2)2 = 0.5
问题15:一个盒子装有14个灯泡,其中5个有缺陷。从盒子中随机抽出3个灯泡,一个接一个地更换。找到有缺陷的灯泡数量的概率分布。
解决方案:
Let X be the random variable denoting the number of defective bulbs drawn in each draw. Since we are drawing a maximum of 3 bulbs at a time, So we can get at max 3 defective bulbs as total defective bulbs are 5.
∴ X can take values 0,1,2 and 3
P(X=0) = P(drawing no defective bulbs)
As we are finding probability for 0 defective bulbs ,so we will select all 3 bulbs
from 9 good bulbs.
n(s) = total possible ways = 14C3
∴ P(X=0) = 9C3/14C3 = 3/13
P(X=1) = P(drawing 1 defective bulbs and 2 good bulbs)
As we are finding probability for 1 defective bulbs ,so we will select 2 bulbs
from 9 good bulbs and 1 from 5 defective ones
∴ P(X=1) = (9C2 x 5C1)/ 14C3 = 45/91
Similarly,
P(X=2) = (9C1 x 5C2)/14C3 = 45/182
P(X=3) = 5C3 /14C3 = 5/182
So, Probability distribution is given below:
xi | pi |
0 | 3/13 |
1 | 45/91 |
2 | 45/182 |
3 | 5/182 |
问题16:在图32.2的轮盘中,轮盘上有13个数字0,1,2,…。,12,这些数字标记在等距的插槽上。玩家在给定的数字上设置₹10。如果球在此插槽中静止,他将从比赛组织者处获得₹100的奖励;否则,他什么也得不到。如果X表示玩家的净损益,请找到E(X)。
解决方案:
As player sets Rs 10 on a number ,if he wins he get Rs 100
∴ his profit is Rs 90.
If he loses, he suffers a loss of Rs 10
He gets a profit when ball comes to rest in his selected slot.
Total possible outcome = 13
Favourable outcomes = 1
∴ probability of getting profit = 1/13
And probability of loss = 12/13
If X is the random variable denoting gain and loss of player
∴ X can take values 90 and -10
P(X=90) = 1/13
And P(X=-10) = 12/13
Now we have pi and xi.
Let’s proceed to find mean
Mean of any probability distribution is given by Mean = ∑xipi
∴ first we need to find the products i.e. pixi and add them to get mean
Following table representing probability distribution gives the required products :
xi | pi | xipi |
90 | 1/13 | 90/13 |
-10 | 12/13 | -120/13 |
E(X) = Mean = 90/13 + (-120/13) = 90/13 – 120/13 = -30/13
问题17:从洗净的52张纸牌中随机抽取三张纸牌(无需更换)。查找红牌数量的概率分布。因此找到分布的均值。
解决方案:
We have total 26 red cards in a deck of 52 cards.
As we are drawing maximum 3 cards at a time so we can get maximum 3 red cards.
If X denotes the number of red cards ,then X can take values from 0,1,2 and 3
P(X=0) = probability of drawing no red cards
We need to select all 3 cards from remaining 26 cards
Total possible ways of selecting 3 cards = 52C3
∴ P(X=0) = 26C3 / 52C3 = 2/17
P(X=1) = P(selecting one red and 2 black cards) = (26C1 x 26C2 ) / 52C3 = 13/34
P(X=2) = P(selecting 2 red and 1 black cards) = (26C2 x 26C1) / 52C3 = 13/34
P(X=3) = 26C3/52C3 = 2/17
So, Probability distribution is given below:
xi | pi | xipi |
0 | 2/17 | 0 |
1 | 13/34 | 13/34 |
2 | 13/34 | 26/34 |
3 | 2/17 | 6/17 |
Mean = 0 + 13/34 + 26/34 + 6/17 = 1.5
问题18:包含5个红色的2黑色的球。随机抽取两个球,无需替换。令X代表绘制的黑色球的数量。 X的可能值是什么? X是随机变量吗?如果是,找到X的均值和方差。
解决方案:
X represents the number of black balls drawn.
∴ X can take values 0,1 and 2
∵ there are total 7 balls
n(S) = total possible ways of selecting 2 balls = 7C2
P(X=0) = P(selecting no black balls) = 5C2/7C2 = 10/21
P(X=1) = P(selecting 1 black ball and 1 red ball)
= (5C1 x 2C1) / 7C2 = 10/21
P(X=2) = P(selecting 2 black ball and 0 red ball) = (5C0 x 2C2) / 7C2 = 1/21
X is said to be a random variable if some of the probabilities associated with each value of X is 1
Here,
P(X=0) + P(X=1) + P(X=2) = 20/42 + 20/42 + 2/42 = 1
∴ X is a random variable.
Now we have pi and xi.
Let’s proceed to find mean and variance.
Mean of any probability distribution is given by Mean = ∑xipi
Variance is given by:
Variance = ∑ xi2pi – (∑xipi)2
∴ first we need to find the products i.e. pixi and pixi2 and add them to get mean and apply the above formula to get the variance.
Following table representing probability distribution gives the required products :
xi | pi | xipi | xi2pi |
0 | 10/21 | 0 | 0 |
1 | 10/21 | 10/21 | 10/21 |
2 | 1/21 | 2/21 | 4/21 |
∴ Mean = 10/21 + 2/21 = 4/7
∴ Variance = 0 + 10/21 + 4/21 – (4/7)2 = 50/147
问题19:从正整数2、3、4、5、6和7中随机选择(不替换)两个数字。X表示获得的两个数字中较大的一个。求出X的概率分布的均值和方差。
解决方案:
∵ two numbers are selected at random like {(2,3) or (5,4) or (4,5)..etc}
Total ways of selecting two numbers without replacement = 6 x 5 =30
As X denote the larger of two numbers selected
∴ X can take values 3,4,5,6 and 7
P(X=3) = P(larger number is 3) = (2/30)[{2,3},{3,2}]
P(X=4) = P(larger number is 4) = (4/30)[{2,4},{4,2},{3,4},{4,3}]
P(X=5) = P(larger number is 5) = (6/30)[{2,5},{3,5},{4,5} and their reverse order]
P(X=6) = P(larger number is 6) = (8/30)[{2,6},{3,6},{4,6},{5,6} and their reverse order]
P(X=7) = P(larger number is 7) = (10/30)[{2,7},{3,7},{4,7},{5,7},{6,7} and their reverse order]
Now we have pi and xi.
Let’s proceed to find mean and variance.
Mean of any probability distribution is given by Mean = ∑xipi
Variance is given by:
Variance = ∑ xi2pi – (∑xipi)2
∴ first we need to find the products i.e. pixi and pixi2 and add them to get mean and apply the above formula to get the variance.
Following table representing probability distribution gives the required products:
xi | pi | xipi | xi2pi |
3 | 2/30 | 6/30 | 18/30 |
4 | 4/30 | 16/30 | 64/30 |
5 | 6/30 | 30/30 | 150/30 |
6 | 8/30 | 48/30 | 288/30 |
7 | 10/30 | 70/30 | 490/30 |
∴ Mean = 6/30 + 16/30 + 30/30 + 48/30 + 70/30 = 17/3
∴ Variance = 18/30 + 64/30 + 150/30 + 288/30 + 490/30 – (17/3)2 = 14/9
问题20:在游戏中,一个人赢得数字大于4会赢取₹5,否则,如果抛出合理的死球,便会失去₹1。该名男子决定掷三次骰子,但当他获得大于4的数字时退出。找到他赢/输的金额的期望值。
解决方案:
We are asked to find the expected amount he wins or lose i.e we have to find the mean of probability distribution of random variable X denoting the win/loss.
As he decided to throw the dice thrice but to quit at the instant he loses
∴ if he wins in all throw he can make earning of Rs 15
If he wins in first two throw and lose in last, he earns Rs (10-1) = Rs 9
If he wins in first throw and then loses ,he earns = Rs 4
If he loses in first throw itself, he earns Rs = -1
Thus X can take values -1,4,9 and 15
P(getting a number greater than 4 in a throw of die) = 2/6 = 1/3
P(getting a number not greater than 4 in a throw of die) = 4/6 = 2/3
P(X=-1) = P(getting number less than or equal to 4) = 2/3
P(X=4) = P(getting > 4) x P(getting ≤ 4) = 1/3 x 2/3 = 2/9
P(X=9) = P(getting > 4) x P(getting > 4) x P(getting ≤ 4) = 1/3 x 1/3 x 2/3 = 2/27
P(X=15) = P(getting > 4) x P(getting > 4) x P(getting > 4) = 1/3 x 1/3 x 1/3 = 1/27
So, Probability distribution is given below:
xi | pi | xipi |
-1 | 2/3 | -2/3 |
4 | 2/9 | 8/9 |
9 | 2/27 | 18/27 |
15 | 1/27 | 15/27 |
∵ Mean = ∑xipi
Mean = -2/3 + 8/9 + 18/27 + 15/27 = 39/27 = 1.44
He can win around Rs 1.45