Probability Practice Questions with Answers

Los ejercicios

1. Ejercicio 1Básico1 puntos

   A fair six-sided die is rolled once. Find the probability of rolling a 4.

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   1/6

   1. 1The sample space is 1, 2, 3, 4, 5, 6 - six equally likely outcomes.
   2. 2Exactly one of them is a 4.
   3. 3P(4) = 1/6
2. Ejercicio 2Básico2 puntos

   A fair six-sided die is rolled once. Find the probability of rolling an even number.

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   1/2

   1. 1Sample space: 1, 2, 3, 4, 5, 6.
   2. 2The even outcomes are 2, 4 and 6 - three of the six.
   3. 3P(even) = 3/6, which simplifies to 1/2.
3. Ejercicio 3Básico2 puntos

   A fair six-sided die is rolled once. Find the probability of not rolling a 6.

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   5/6

   1. 1P(6) = 1/6.
   2. 2Not rolling a 6 is the complement, and the two probabilities must total 1.
   3. 3P(not 6) = 1 - 1/6 = 5/6
   4. 4Counting directly agrees: 1, 2, 3, 4 and 5 are five outcomes out of six.
4. Ejercicio 4Básico3 puntos

   A bag holds 5 red, 3 blue and 4 green counters. One counter is taken at random. Find P(blue) and P(not blue).

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   P(blue) = 1/4 and P(not blue) = 3/4

   1. 1Total counters: 5 + 3 + 4 = 12.
   2. 2P(blue) = 3/12, which simplifies to 1/4.
   3. 3P(not blue) = 1 - 1/4 = 3/4. Counting directly: 5 + 4 = 9 counters are not blue, and 9/12 = 3/4.
   4. 4Check: 1/4 + 3/4 = 1.
5. Ejercicio 5Estándar2 puntos

   A spinner has red, blue and green sections only. P(red) = 0.35 and P(blue) = 0.2. Find P(green).

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   0.45

   1. 1The spinner must land on one of the three colours, so the three probabilities add to 1.
   2. 20.35 + 0.2 = 0.55
   3. 3P(green) = 1 - 0.55 = 0.45
6. Ejercicio 6Estándar3 puntos

   Two fair coins are tossed. Find the probability of exactly one head, and the probability of at least one head.

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   P(exactly one head) = 1/2; P(at least one head) = 3/4

   1. 1Write the sample space: HH, HT, TH, TT - four equally likely outcomes.
   2. 2Exactly one head happens in HT and TH: 2 out of 4, so 1/2.
   3. 3At least one head covers HH, HT and TH: 3 out of 4.
   4. 4Treating this as three outcomes - two heads, one head, two tails - gives 1/3 and is wrong, because HT and TH are different.
7. Ejercicio 7Estándar3 puntos

   Two fair six-sided dice are rolled and the scores are added. Find the probability that the total is 7.

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   1/6

   1. 1The sample space has 6 × 6 = 36 equally likely outcomes.
   2. 2Totals of 7 come from (1,6), (2,5), (3,4), (4,3), (5,2) and (6,1) - six outcomes.
   3. 3P(total 7) = 6/36 = 1/6
   4. 47 has more combinations than any other total, which is why it is the most likely score.
8. Ejercicio 8Estándar3 puntos

   Two fair six-sided dice are rolled and the scores are added. Find the probability that the total is 10.

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   1/12

   1. 1The sample space is still 36 outcomes.
   2. 2Totals of 10 come from (4,6), (5,5) and (6,4) - three outcomes.
   3. 3(5,5) is one outcome, not two: there is only one way for both dice to show 5.
   4. 4P(total 10) = 3/36 = 1/12
9. Ejercicio 9Estándar2 puntos

   A drawing pin is dropped 200 times and lands point up 130 times. Estimate the probability that it lands point up.

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   13/20, or 0.65

   1. 1A drawing pin is not symmetrical, so the two outcomes are not equally likely and the probability cannot be reasoned out. It has to be estimated from the data.
   2. 2Relative frequency = number of successes ÷ number of trials = 130 ÷ 200.
   3. 3130/200 = 13/20 = 0.65
   4. 4So P(point up) is about 0.65 and P(point down) about 0.35. More trials would sharpen the estimate; nothing makes 0.5 correct.
10. Ejercicio 10Avanzado3 puntos

    A bag holds 4 red and 6 yellow counters. A counter is taken, its colour noted, and it is put back. A second counter is then taken. Find the probability that both are red.

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    4/25

    1. 1There are 10 counters, and the first is replaced, so the second draw faces exactly the same bag.
    2. 2P(red) = 4/10 on each draw.
    3. 3Both red means the first red and the second red, so multiply: 4/10 × 4/10 = 16/100.
    4. 416/100 simplifies to 4/25.
11. Ejercicio 11Avanzado3 puntos

    A bag holds 4 red and 6 yellow counters. Two counters are taken without replacement. Find the probability that both are red.

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    2/15

    1. 1First draw: 4 red out of 10 counters, so 4/10.
    2. 2One red counter has gone, so the second draw is from 9 counters of which 3 are red: 3/9.
    3. 3Multiply along the branch: 4/10 × 3/9 = 12/90.
    4. 412/90 simplifies to 2/15. Both numbers change - keeping the denominator at 10 is the usual error.
12. Ejercicio 12Avanzado4 puntos

    A bag holds 4 red and 6 yellow counters. Two counters are taken without replacement. Find the probability that one is red and one is yellow.

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    8/15

    1. 1One of each can happen two ways: red then yellow, or yellow then red. Both count.
    2. 2P(red then yellow) = 4/10 × 6/9 = 24/90.
    3. 3P(yellow then red) = 6/10 × 4/9 = 24/90.
    4. 4The two branches are separate routes to the same event, so add: 24/90 + 24/90 = 48/90 = 8/15.
    5. 5Check: P(both red) = 12/90 and P(both yellow) = 6/10 × 5/9 = 30/90, and 12 + 30 + 48 = 90, so the three total 1.

Write the sample space out, even when it feels unnecessary

Almost every probability mistake at this level is a counting mistake, and almost every counting mistake disappears when the outcomes are listed. Two coins have four outcomes, not three: HH, HT, TH and TT. Two dice have thirty-six, not eleven. Writing them down takes half a minute and removes the guesswork entirely.

The word equally likely is doing quiet work in all of this. A probability is the number of outcomes you want divided by the total only when every outcome in that list has the same chance. HT and TH are two separate outcomes, which is precisely why exactly one head has probability 1/2 rather than 1/3.

The check that takes five seconds

Every possible outcome of an experiment has to happen exactly once, so the probabilities of a complete set of outcomes always add to 1. That gives a free check on almost every answer, and it also solves whole questions on its own: the spinner question below is answered by subtraction and nothing else.

It is worth applying at the end of the two-counter questions too. Drawing two counters from a bag of four red and six yellow gives three outcomes: both red, both yellow, one of each. Their probabilities come to 12/90, 30/90 and 48/90, which add to exactly 1. If they do not add to 1, one of the three is wrong and you know it before checking the answers.

With replacement, and without

If the counter goes back in the bag, the second draw faces the identical situation as the first, and both fractions have the same denominator. If it does not go back, both numbers change: one fewer counter of that colour, and one fewer counter altogether.

So 4/10 × 4/10 becomes 4/10 × 3/9. The denominator falling from 10 to 9 is the part most often missed, because the numerator change feels like the whole of the adjustment. Read the question for the words replaced, put back or without replacement before writing anything, since nothing else in the working distinguishes the two cases.