Counter example. The sum of three fractions with odd numerators is never 1/2

Mathematics

2 answers:

kolbaska11 [484]2 years ago

8 0

1/6 + 1/6 + 1/6 = 3/6 = 1/2

tia_tia [17]2 years ago

5 0

One <em>possible counterexample</em> is:

1/4 + 1/5 + 1/20 = 1/2.

In this problem, we would find the least common denominator for 4, 5 and 20. The first thing all 3 numbers will divide into is 20:

1/4 = 5/20; 1/5 = 4/20; 1/20 = 1/20

This gives us: 5/20 + 4/20 + 1/20

Adding the numerators, we get (5+4+1)/20 = 10/20 = 1/2.

You might be interested in

A bin of 5 transistors is known to contain 2 that are defective. The transistors are to be tested, one at a time, until the defe

xxMikexx [17]

Answer:

$P(N_1 = a , N_2 = b)= \frac{1}{5-a C 1} * \frac{5-a C 1}{5C2} = \frac{1}{5C2}=\frac{1}{10}$

Step-by-step explanation:

For the random variable $N_1$ we define the possible values for this variable on this case $[1,2,3,4,5]$ . We know that we have 2 defective transistors so then we have 5C2 (where C means combinatory) ways to select or permute the transistors in order to detect the first defective:

$5C2 = \frac{5!}{2! (5-2)!}= \frac{5*4*3!}{2! 3!}= \frac{5*4}{2*1}=10$

We want the first detective transistor on the ath place, so then the first a-1 places are non defective transistors, so then we can define the probability for the random variable $N_1$ like this:

$P(N_1 = a) = \frac{5-a C 1}{5C2}$

For the distribution of $N_2$ we need to take in count that we are finding a conditional distribution. $N_2$ given $N_1 =a$ , for this case we see that $N_2 \in [1,2,...,5-a]$ , so then exist $5-a C 1$ ways to reorder the remaining transistors. And if we want b additional steps to obtain a second defective transistor we have the following probability defined: