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TOPIC: Probability
#2454
Sreeks (User)
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Re:Probability 1 Year ago Karma: 0  
Here i go...

1. 1/4
2.0.0001
4.4782969
5.900

I am not clear with 3rd one...
 
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#2464
Vijay (User)
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Re:Probability 1 Year ago Karma: 2  
bhavin wrote:
P.S : Vijay, i have copied your harder probability questions here :

Everybody, pls post all your probability related queries here !!

Here are the questions from Vijay :

A 1-inch-diameter coin is throw n on a table covered wit h a grid of lines tw o inches apart. What is the probabilit y t he coin lands in a square wit hout touching any of the lines of t he grid?

1/4
1/2
1/3
1/6
1/8


An airplane is built to be able to f ly on one engine. If the plane' s two engines operate independently, and each has a 1% chance of failing in any given f our-hour flight , w hat is the chance t he plane w ill fail t o complete a four-hour flight to Oklahoma due t o engine f ailure?

.0001
.001
.00001
.01
1


In a roomful of 30 people, what is the probabilit y t hat at least two people have the same birt hday? Assume birt hdays are uniformly dist ributed and there is no leap year complication.

70
60
50
35
80


How many 7-digit numbers have no two adjacent digits equal?

4,782,969
5,782,969
4,600,969
4,782,600
4,782,999


How many 5-digit palindromes are there? (A palindrome is a number that reads the same way forwards and backwards. For example, 27872 and 48484 are palindromes, but 28389 and 12541 are not.)

900
1000
784
512
625



Correct answers are:

1/4

i AM NOT SURE ABOUT 2ND ANSWER.In the notes i am having its 0.001, but i need to know the approach.

70
 
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#2465
Vijay (User)
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Re:Probability 1 Year ago Karma: 2  
Answer for question 4

We have 9 choices for the first digit (since it cannot be 0). Then, no matter what we choose for the first digit, we have 9 choices for the second digit—any digit except the one that we chose for the first digit. Similarly, for each subsequent digit, we have 9 choices, since each digit can be any of 0 through 9 as long as it does not match the previous digit. Therefore, there are 9 choices for every digit, and hence there are 97 = 4,782,969 such numbers.
 
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#2466
Vijay (User)
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Re:Probability 1 Year ago Karma: 2  
Answer for question 5:

We have 9 choices for the first digit (since it can’t be 0), then 10 choices for the second digit, then 10 choices for the third digit. But now
we’re out of choices—the fourth digit must match the second, and the last digit must match the first.

Therefore, there are 9 · 10 · 10 = 900 such numbers.
 
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#2506
bhavin (Moderator)
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Re:Probability 1 Year ago Karma: 25  
bhavin wrote:


An airplane is built to be able to f ly on one engine. If the plane' s two engines operate independently, and each has a 1% chance of failing in any given f our-hour flight , w hat is the chance t he plane w ill fail t o complete a four-hour flight to Oklahoma due t o engine f ailure?

.0001
.001
.00001
.01
1



Good explanations for the last 2 sums vijay !!

I do not concur with the answer option for the second question i.e the quoted question !!

In my opinion , correct answer should be 0.0001 ...Ans option A
airplane fails only if both the engine fail !!

And for independent event, p(A& = p(A) * p(

Hence, p (failure) = p(failure of both engines) = 0.01 * 0.01 = 0.0001 ...Ans A

Cheers !!
 
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#2507
bhavin (Moderator)
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Re:Probability 1 Year ago Karma: 25  
bhavin wrote:

In a roomful of 30 people, what is the probabilit y t hat at least two people have the same birt hday? Assume birt hdays are uniformly dist ributed and there is no leap year complication.

70
60
50
35
80



Vijay, i still have too many reservations for the above problem ...

First of all, answer options for a probability question can never be integral values above 1 ..It has to be a fraction between 0 and 1, both inclusive ...

Secondly, Question is Ambiguous
 
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