1. The probability of winning a certain game is 1/4. How many times must the game be played for there to be be at least a 50% chance of winning at least once ?
2; 3; 4; 5 or 6 times.

The smallest answer is 2, so let's start with 2 games:

The only outcome we don't want is 2 losses. The chance of 2 losses is 3/4 * 3/4 = 9/16. Therefore, the chance of at least 1 win is 1 - 9/16 = 7/16. Not a half, so (a) is wrong.

At this point 3 is looking like a great guess, since we're already so close to 1/2. If you're short on time, guess (B). But, let's work it out.

If we play 3 games and want to win at least 1, then the only outcome we don't want is 3 losses. The chance of 3 losses is 3/4 * 3/4 * 3/4 = 27/64. Therefore, the chance of at least 1 win is 1 - 27/64 = 37/64 which is > 1/2. Therefore, (B) is correct.

2. If one positive single digit even number and one positive single digit odd number are chosen at random, and their sum is 11, what is the probability that the odd number is equal to 9 ?
1/10; 1/5; 1/4; 1/3 or 1/2

There are 4 ways for a pair of single digit numbers to add to 11: 2+9, 3+8, 4+7 and 5+6.

Therefore, there's a 1/4 chance that 9 is in the pair.

3. If a coin is flipped five times, what is the probability that you get either heads or tails twince in a row during five flips ?
1/32; 1/16; 1/8; 5/32 or 3/5

Are you sure that you copied the question correctly? The correct answer to the question you posted isn't among those choices. You're missing either "DO NOT get either heads or tails twice in a row" or "get either heads or tails EXACTLY twice in a row".

As you've posted it, the answer should be 15/16, since there are only two sequences of flips where we DO NOT get heads or tails twice in a row: HTHTH and THTHT. Since there are 32 possible strings (2^5) for 5 flips, that means that we don't want 2/32 or 1/16 possibilities, which means that we DO want 1 - 1/16 = 15/16 possibilities.

4. On a multiple choice test with 5 questions and three answer choices per question, what is the probability of answering at least four questions correctly if each answer is chosen randomly ?
5/243; 10/243; 11/243; 1/15 or 1/12

If we want to answer at least 4 questions correctly, then there are 2 scenarios taht we want: 4/5 and 5/5.

The chance of getting 4/5 is (1/3)^4 * (2/3)^1 = 2/243. Since we don't care which 4 we get right, there are 5C4 (5 choose 4) = 5 different ways we can get 4/5 correct. Therefore, the chance of getting 4/5 correct is 5 * 2/243 = 10/243.

The chance of getting all 5 right is (1/3)^5 = 1/243.

So, the chance of getting AT LEAST 4/5 right is 10/243 + 1/243 = 11/243.

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Hope that helps!