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Chapter 18: Problem 802

Out of 20 consecutive whole numbers two are chosen at random. Then the probability that their sum is odd is (a) \((5 / 19)\) (b) \((10 / 19)\) (c) \((9 / 19)\) (d) \((11 / 19)\).

Short Answer

Expert verified
The probability of the sum of two randomly chosen whole numbers being odd from a set of 20 consecutive whole numbers is \(\frac{10}{19}\).

Step by step solution

01

Identify scenarios for picking two whole numbers

Since the problem states we have 20 consecutive whole numbers, we can think of them as 10 pairs of even and odd numbers. The possible scenarios include picking an odd number and then an even number (OE), picking an even number and then an odd number (EO), picking two even numbers (EE), or picking two odd numbers (OO). Since the sum of an even and an odd number is always odd, we are interested in the probability of choosing OE or EO pairs.
02

Determine the total number of ways to pick 2 numbers

In order to calculate the probability, we'll need to determine the total number of ways to pick two numbers from the 20 consecutive whole numbers set. This can be done using combinations, as it doesn't matter which order we choose the two numbers in. Using the combination formula, we have: \[C(n, k) = \frac{n!}{k!(n-k)!}\] In our case, n = 20 and k = 2, so we can calculate the total combinations: \[C(20, 2) = \frac{20!}{2!(20-2)!} = \frac{20!}{2!18!} = \frac{20 \times 19}{2} = 190\]
03

Calculate the number of OE and EO pairs

We have 10 odd numbers and 10 even numbers in our set. To calculate the number of OE and EO pairs, we can multiply the number of odd numbers by the number of even numbers: \[OE = 10 \times 10 = 100\] \[EO = 10 \times 10 = 100\] The total number of OE and EO pairs is the sum of these two values: \[OE + EO = 100 + 100 = 200\] However, we have double-counted each combination because the order doesn't matter in our probability calculation (OE and EO are considered the same). Therefore, we need to divide the result by 2: \[\frac{200}{2} = 100\]
04

Calculate the probability of the sum being odd

Now that we have the total number of favorable outcomes (OE and EO pairs) and the total number of possible outcomes (all pairs), we can calculate the probability of choosing two numbers with an odd sum: \[P(\text{odd sum}) = \frac{\text{favorable outcomes}}{\text{total outcomes}} = \frac{100}{190} = \frac{10}{19}\]
05

Choose the correct answer

Our calculated probability of the sum of two chosen numbers being odd is \(\frac{10}{19}\). Comparing this to the given answer choices, we can see that the correct option is: (b) \(\frac{10}{19}\)

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Most popular questions from this chapter

A dice is loaded so that the probability of face \(\mathrm{i}\) is proportional to i. \(\mathrm{i}=1,2, \ldots .6\). Then the probability of an even number occupy when the dice is rolled is (a) \((2 / 7)\) (b) \((3 / 7)\) (c) \((4 / 7)\) (d) \((5 / 7)\)

If coefficient of variation \(=70\) and mean \(=10\) then variance is (a) 49 (b) 7 (c) 100 (d) 80

A die is thrown 3 times and the sum of the thrown numbers is 15 . The probability for which the number 5 appears in first throw is (a) \((3 / 10)\) (b) \((1 / 36)\) (c) \((1 / 9)\) (d) \((1 / 3)\)

Find mean and S.D. from given data $$ \begin{array}{|l|c|c|c|c|c|} \hline \text { Class } & 33-35 & 36-38 & 39-41 & 42-44 & 45-47 \\ \hline \text { Frequency } \mathrm{f} & 17 & 19 & 23 & 21 & 20 \\ \hline \end{array} $$ (a) \(40.24,4.20\) (b) \(40.24,4.30\) (c) \(4.5,40.20\) (d) \(40.24,4.90\)

Variance of \(1,3,5,7 \ldots \ldots \ldots(4 n+1)\) is (a) \([\\{2 n(2 n-1)\\} / 3]\) (c) \((1 / n) \sqrt\left[\left(n^{2}-1\right) / 3\right] 100\) (d) \([\\{4 n(n+1)\\} / 3]\)
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