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QuestionWhat is the probability that the sum of four randomly chosen numbers from {1,2,3,…,10}\{1,2,3,\ldots,10\} is odd?

Studdy Solution

STEP 1

Assumptions1. The set of numbers to choose from is {1,,3,…,10}\{1,,3, \ldots,10\} . Four numbers will be randomly chosen from this set3. We are looking for the probability that the sum of these four numbers will be an odd number

STEP 2

First, we need to understand that an odd number can only be obtained by adding an odd number of odd numbers or an even number of even numbers. In our case, since we are choosing four numbers, we can have three cases1. All four numbers are odd2. Two numbers are odd and two are even. All four numbers are even

STEP 3

We know that there are5 odd numbers and5 even numbers in the set. So, the total number of ways to choose numbers from the set of10 numbers is given by the combination formulaTotal ways=C(10,)Total\, ways = C(10,)

STEP 4

Now, let's calculate the total number of ways to choose4 numbers from the set of10 numbers.
Total ways=C(10,4)=10!4!(10−4)!=210Total\, ways = C(10,4) = \frac{10!}{4!(10-4)!} =210

STEP 5

Next, we calculate the number of ways to choose4 odd numbers from the set. This is given by the combination formulaOdd ways=C(5,4)Odd\, ways = C(5,4)

STEP 6

Calculate the number of ways to choose4 odd numbers from the set.
Odd ways=C(5,4)=5!4!(5−4)!=5Odd\, ways = C(5,4) = \frac{5!}{4!(5-4)!} =5

STEP 7

Next, we calculate the number of ways to choose2 odd and2 even numbers from the set. This is given by the product of the combination of choosing2 odd numbers from5 odd numbers and choosing2 even numbers from5 even numbersEven ways=C(5,2)×C(5,2)Even\, ways = C(5,2) \times C(5,2)

STEP 8

Calculate the number of ways to choose2 odd and2 even numbers from the set.
Even ways=C(5,2)×C(5,2)=5!2!(5−2)!×5!2!(5−2)!=10×10=100Even\, ways = C(5,2) \times C(5,2) = \frac{5!}{2!(5-2)!} \times \frac{5!}{2!(5-2)!} =10 \times10 =100

STEP 9

The case where all four numbers are even is not possible because the sum of four even numbers is always even. So, the total number of favorable ways is the sum of the number of ways to choose4 odd numbers and the number of ways to choose2 odd and2 even numbers.
avorable ways=Odd ways+Even waysavorable\, ways = Odd\, ways + Even\, ways

STEP 10

Calculate the total number of favorable ways.
avorable ways=Odd ways+Even ways=5+100=105avorable\, ways = Odd\, ways + Even\, ways =5 +100 =105

STEP 11

Finally, the probability that the sum of four numbers chosen will be an odd number is given by the ratio of the number of favorable ways to the total number of ways.
Probability=avorable waysTotal waysProbability = \frac{avorable\, ways}{Total\, ways}

STEP 12

Calculate the probability.
Probability=avorable waysTotal ways=105210=0.5Probability = \frac{avorable\, ways}{Total\, ways} = \frac{105}{210} =0.5So, the probability that the sum of four numbers chosen will be an odd number is0.5.

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