Conditional probability questions ?

In summary, conditional probability is the likelihood of an event occurring given that another event has already occurred. It differs from regular probability in that it takes into account the occurrence of another event and is calculated based on a subset of all possible outcomes. To calculate conditional probability, divide the probability of the joint occurrence of two events by the probability of the first event. It has various real-world applications in fields such as statistics, finance, economics, and psychology. To improve understanding, practice solving different types of problems and familiarize yourself with basic principles and formulas, as well as reading about real-world examples.
  • #1
ngkamsengpeter
195
0
Please help me to solve the following questions :
1) There are three box : box X has 10 bulbs which 4 are defective
Box Y has 6 bulbs which 1 are defective
Box Z has 8 bulbs which 3 are defective
a box is chosen at random and the buld is randomly selected from the choosen box . If the bld is not defective, find the probabily that it come from box Z .(Answer = 75/247)

2) A pair of fair dice is thrown . If two numbers are different , find the probability that
a)the sum is 6( answer=2/15)
b)an ace appears (answer=1/3)
c)the sum is 4 or less (answer = 2/15)
d)the sum is even (answer is 2/5)
the sum exceeds nine (answer is 2/15)

3)One couple has two children , find the probability both children are boys if
a)at least one of the children is a boy (answer is 1/3)
b)the older child is a boy (answer is 1/2)

I cannot solve most of the questions above because I don't know how to find the P(A n B) since P(A|B)=P(A n B)/P(B) . So, please help me to solve the questions .Thank you .
 
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  • #2
ngkamsengpeter said:
Please help me to solve the following questions :
1) There are three box : box X has 10 bulbs which 4 are defective
Box Y has 6 bulbs which 1 are defective
Box Z has 8 bulbs which 3 are defective
a box is chosen at random and the buld is randomly selected from the choosen box . If the bld is not defective, find the probabily that it come from box Z .(Answer = 75/247)

2) A pair of fair dice is thrown . If two numbers are different , find the probability that
a)the sum is 6( answer=2/15)
b)an ace appears (answer=1/3)
c)the sum is 4 or less (answer = 2/15)
d)the sum is even (answer is 2/5)
the sum exceeds nine (answer is 2/15)

3)One couple has two children , find the probability both children are boys if
a)at least one of the children is a boy (answer is 1/3)
b)the older child is a boy (answer is 1/2)

I cannot solve most of the questions above because I don't know how to find the P(A n B) since P(A|B)=P(A n B)/P(B) . So, please help me to solve the questions .Thank you .

If P(A|B)= P(A n B)/P(B) then P(A n B)= P(A|B)P(B)! In particular, if
either event A or B must happen, and A and B are mutually exclusive, then
P(C)= P(C|A)P(A)+ P(C|B)P(B).

Now:
1) There are three box : box X has 10 bulbs which 4 are defective
Box Y has 6 bulbs which 1 are defective
Box Z has 8 bulbs which 3 are defective
a box is chosen at random and the buld is randomly selected from the choosen box . If the bld is not defective, find the probabily that it come from box Z .(Answer = 75/247)
.
Let N be the event that a non-defective bulb is chosen. Then
P(N)= P(N|X)P(X)+ P(N|Y)P(Y)+ P(N|Z)P(Z). We can also see that, since P(N|Z)P(Z)= P(Z|N)P(N),
P(Z|N)= P(N|Z)P(Z)/P(N)= P(N|Z)P(Z)/(P(N|X)P(X)+ P(N|Y)P(Y)+ P(N|Z)P(Z)) where all of the probabilities on the right are "a-priori"- before we select a bulb.
There are 3 boxs and we choose them at random so a-priori P(X)= P(Y)= P(Z)= 1/3. X has 10 bulbs of which 6 are non-defective so P(N|X)= 6/10= 3/5. Y has 6 bulbs of which 5 are non-defective so P(N|Y)= 5/6. Z has 8 bulbs of which 5 are non-defective so P(N|Z)= 5/8. That is,
P(N|X)P(X)+ P(N|Y)P(Y)+ P(N|Z)P(Z)= (3/5)(1/3)+ (5/6)(1/3)+ (5/8)(1/3)= 247/360.
Again, P(N|Z)= 5/8 and P(Z)= 1/3 so P(Z|N)= P(N|Z)P(Z)/(P(N|X)P(X)+ P(N|Y)P(Y)+ P(N|Z)P(Z)) becomes P(Z|N)= (5/8)(1/3)/(247/360)=
(5/24)(360/247)= (5)(15)/247= 75/247.

2) A pair of fair dice is thrown . If two numbers are different , find the probability that
a)the sum is 6( answer=2/15)
b)an ace appears (answer=1/3)
c)the sum is 4 or less (answer = 2/15)
d)the sum is even (answer is 2/5)
the sum exceeds nine (answer is 2/15)
You know that there are 6*6= 36 "equally likely combinations, 6 of which have both numbers the same so 30 have both numbers different. Now,
(a) how many of those add to 6? ((1,5),(2, 4), (4,2), (1,5)).
(b) how many of those have an ace?
(c) how many of those sum to 4 or less?
(d) how many of those have even sum?
(e?) how may of those have sum greater than 9?

3)One couple has two children , find the probability both children are boys if
a)at least one of the children is a boy (answer is 1/3)
b)the older child is a boy (answer is 1/2)
Surprising that those are different answers, isn't it?
Assuming birth of boy or girl is equally likely, there are four equally likely ways to have two children. (Boy, Boy), (Boy, Girl), (Girl, Boy), (Girl, Girl) where they are listed in order of birth.
If "at least one of the children is a boy", the we can ignore (Girl, Girl) and have only (Boy, Boy), (Boy, Girl), (Girl, Boy). From that information, what is the probability of (Boy, Boy)?
If "the older child is a boy, then we can ignore both (Girl, Girl) and (Girl, Boy). Now what is the probability of (Boy, Boy)?
 

What is conditional probability?

Conditional probability is the likelihood of an event occurring given that another event has already occurred. It is calculated by dividing the probability of the joint occurrence of both events by the probability of the first event.

How is conditional probability different from regular probability?

Regular probability deals with the likelihood of a single event occurring, while conditional probability takes into account the occurrence of another event. It is calculated based on a subset of all possible outcomes, rather than the entire sample space.

How do you calculate conditional probability?

Conditional probability is calculated by dividing the probability of the joint occurrence of two events by the probability of the first event. This can be represented by the equation P(A|B) = P(A and B) / P(B).

What are some real-world applications of conditional probability?

Conditional probability is commonly used in fields such as statistics, finance, economics, and psychology. Some real-world applications include predicting stock prices, analyzing consumer behavior, and determining the effectiveness of medical treatments.

How can I improve my understanding of conditional probability?

To improve your understanding of conditional probability, it is helpful to practice solving different types of conditional probability problems and familiarize yourself with the basic principles and formulas. You can also read about real-world applications and examples to see how it is used in different fields.

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