Three numbers are chosen at random without replacement from {1, 2, 3, ..., 10}. What is the probability that the minimum of the chosen number is 3, or their maximum is 7?

Payton George

Payton George

Answered question

2022-10-22

Probability of a number randomly chosen out of given numbers
Three numbers are chosen at random without replacement from {1, 2, 3, ..., 10}. What is the probability that the minimum of the chosen number is 3, or their maximum is 7?
Approach 1:
Let A and B denote the following events
A:minimum of the chosen number is 3
B: maximum of the chosen number is 7
We have,
P ( A ) = P ( choosing 3 and two other numbers from 4 to 10 ) = ( 7 2 ) ( 10 3 ) = ( 7 6 ) 2 3 2 10 9 8 = 7 40
P ( B ) = P ( choosing 7 and two other numbers from 1 to 6 ) = ( 6 2 ) ( 10 3 ) = 6 5 2 3 2 10 9 8 = 1 8
P ( A B ) = P ( Choosing 3 and 7 and one other number from 4 to 6 ) = = 3 ( 10 3 ) = 3 3 2 10 9 8 = 1 40
Now P ( A B ) = P ( A ) + P ( B ) P ( A B )
= 7 40 + 1 8 1 40 = 11 40
Approach 2:
Chance of a number being drawn out of the given 10 numbers is equiprobable. Hence
P(any one number drawn )= 1 10
P(A)= ( 7 2 ) 1 10 7 9 6 8 = 21 42 720 = 14 7 80 = 98 80 which is wrong as it is above 1.
Here, 1 10 is the probability of getting 3 and 7 9 6 8 is the probability of getting two number above 3 which can be chosen in ( 7 2 ) ways.
Why I cannot find the P(A) like this ? Please tell me what is wrong in approach 2

Answer & Explanation

wlanauee

wlanauee

Beginner2022-10-23Added 17 answers

Step 1
Your approach 2 has three errors. First, you are double multiplying by ( 7 2 ) , once when you use the binomial and once in the numerators of 7 9 and 6 8 .
Step 2
When you use 7 9 and 6 8 you are already choosing the numbers. You have a factor 2 when you use 7 9 and 6 8 (note 7 6 = 2 ( 7 2 ) )) because there are two orders to select each pair and you count them both. Finally, you only count cases where 3 is the first number selected.
Nayeli Osborne

Nayeli Osborne

Beginner2022-10-24Added 2 answers

Step 1
The correct working for Approach 2, (general multiplication rule) using your notation:
P ( A ) = P (choosing 3 and two other numbers from 4 10 ) = 3 10 7 9 6 8 = 7 40
P ( B ) = P(choosing 7 and two other numbers from 1 6 ) = 3 10 6 9 5 8 = 1 8
Step 2
P ( A B ) = P(Choosing 3 and 7 and one other number from 4 6 ) = 3 10 3 9 2 8 = 1 40
The figures now tally, and you can proceed further as before

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