## How much evidence is enough evidence?

**FACT:** Each card has a number on one side and a letter on the other.

**HYPOTHESIS: **Every card that has a vowel on one side has an even number on its opposite side.

**THE TEST:** Which card or cards must you turn over in order to prove (or disprove) the Hypothesis? Which is easier to do: prove or disprove?

(Obviously, you COULD turn over all four cards, and perhaps you must. But if you can do it with fewer, you’ll save yourself time and energy gathering your evidence. Try for fewer than four.)

Type your answer in the *Reply *field below and Save.

The 15 possible answers are:

**All four cards:** G 2 3 E**Three cards:** G 2 3 / G 2 E / G 3 E / 2 3 E**Just two cards:** G and 2 / G and 3 / G and E / 2 and 3 / 2 and E / 3 and E**Just one card:** Card G only / Card 2 only / Card 3 only / Card E only

2 and E

I would turn over G 2 3 / G 2 E / G 3 E / 2 3 E since these cards have the most amount of vowels.

Card E only

In order to test this hypothesis you would have to flip G and 2 , in order to see if 2 contains a vowel and if G is an odd number.

It is impossible (in my uneducated, initial opinion) to confirm or deny the hypothesis without turning over all of the cards. If we turn over 1 card, it’s possible to confirm the hypothesis, but it’s impossible to know for sure unless we check all of them. If the single card we first turn over does not meet the requirements of the hypothesis, then we only have to turn 1 over. Therefore, we must turn over (infinity) amount of cards and every single one of those cards must fall in line with the hypothesis or it will be denied. Once it is denied, we can be certain about the nature of the hypothesis; that it is wrong. If it is never denied, we can still never be certain because the next card might be the one that falls out of line.

you would only need to turn over the 2 and the E. Only because there is no constraint on what is on the back of the vowels or odd numbers. Therefore, anything could be on the back. You can test the hypothesis just by flipping the 2 and the E

*Sorry I meant consonants not vowels

The cards you must turn over are cards G and 3.

The cards that I would turn over in order to test the hypothesis would be the 2 and 3 cards because there may be a word that could be uncovered.

– 2&E

Process of elimination

I would turn over cards 2 & E only

Explanation more in notes*

All four cards would need to be flipped over.

The only card you need to turn over is card E because the Hypothesis states that every card that has a vowel on one side has an even number on the other side, meaning that if you see a vowel, that means the other side has an even number. However, this does not mean that if a card has an even number, it has to have a vowel on the other side of it. To prove the hypothesis we just need to flip over any cards that have a vowel on it.

I would turn over 2 and E because 2 is an even number and e is a vowel.

I would turn over G23/G2E/G3E/23E because the cards have a balance of numbers and letters

I would flip car 2 and card E because one is a even number and the other is a vowel.

I need 2,3,E to prove or disprove this hypothesis.

I would turn over card 2 and card E.

You could turn over just the card E and see if it has an even number then we would know if the hypothesis is true or not. It is easier to disprove the hypothesis.

2 and E

23E

I would need only up to 3 cards. First E, to see if the other side is even. Then 2, to see if the other side is a vowel. If both of those prove the hypothesis true, then I would flip 3, to see if the other side is a vowel, which will give the definitive answer of whether the hypothesis is true or untrue.

2 and E if the hypothesis is true there should be an even number on the back of E and a vowel on the back of 2

turn over 2 and e

2 cards would need to be flipped, e and 2. E because it is the only vowel and 2 since it is the only even. This may not prove the hypothesis but it is the only two that have a chance at it

flip over 2 and E

I would turn over card G, 2 and E

Card E Only

G and 3

Turn over E and 2. Considering 2 is an even number and E is a vowel. Also it appeals the eye better for the first two cards having letters and the last two being numbers.

I would flip all cards to give me either the best chance to disprove or gather the most about the evidence.

2 and E

You would only need to flip card E over because every card with a vowel on one side has an even number but that does not mean that every card that has an even number has to have a vowel on the other side of it.

I would turn over the E and 3

G2E

2&E

Turning over 2 & E would give you the answer you need

I would turn over 2 and E to prove the hypothesis

2 and E

In order to test this hypothesis, I would personally turn over card E and card G. Because we’re testing to see if there is an even number behind the vowel, and also seeing if there’s a possibility there’s also an even number behind a card that is not a vowel.

I would turn over card E

I would turn over 2 and E.

Flipping both E and 3 will prove it or disprove it definitively. Flipping 2 could disprove it but there is no guarantee that it will.

I would turn over the 2 and the E

2 and E.

Given that we only need to know about the vowel and/or even numbered cards, the other cards become irrelevant to check.

The only way to know for sure is to flip over 2 and E. Both G and 3 do not matter at all, since G isn’t a vowel and 3 isn’t an even number. I believe students are most likely to try and confirm, rather than disprove.

I got confused haha

I would flip both E and 2 since E according the hypothesis would be an even number and 2 is already an even number then the other side of the card should also be a vowel according to the hypothesis.

By flipping over card E and card 3 will help find the slut ion too this problem because by flipping over E we can prove or disprove and aspect of our hypothesis off the bat but by also flipping over card 3 we can also prove the extent of the hypothesis compared to other factors.

I would turn 2 and E only

G 2 3 E