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Probability

17 practice questions with full worked solutions. Free, no account needed.

Problems & worked solutions

Problem #0463 International

Problem 1 — Probability

A card is drawn at random from a standard 5252-card deck. The probability that it is a heart is:

Show answer & worked solution
  1. A. 152\dfrac{1}{52}
  2. B. 113\dfrac{1}{13}
  3. C. 14\dfrac{1}{4}✓ correct
  4. D. 12\dfrac{1}{2}

The deck has 1313 hearts out of 5252 cards: 1352=14\dfrac{13}{52} = \dfrac{1}{4}.

Problem #0461 International

Problem 2 — Probability

A fair coin is tossed once. The probability of obtaining heads is:

Show answer & worked solution
  1. A. 00
  2. B. 12\dfrac{1}{2}✓ correct
  3. C. 11
  4. D. 14\dfrac{1}{4}

A fair coin has two equally likely outcomes; the probability of heads is 12\dfrac{1}{2}.

Problem #0462 International

Problem 3 — Probability

A fair 66-sided die is rolled. The probability of obtaining an even number is:

Show answer & worked solution
  1. A. 13\dfrac{1}{3}
  2. B. 12\dfrac{1}{2}✓ correct
  3. C. 23\dfrac{2}{3}
  4. D. 16\dfrac{1}{6}

Even outcomes: {2,4,6}\{2, 4, 6\}, three out of six. Probability =36=12= \dfrac{3}{6} = \dfrac{1}{2}.

Problem #0947 US SAT

Problem 4 — Classical Probability

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

Show answer & worked solution
  1. A. 13\dfrac{1}{3}
  2. B. 12\dfrac{1}{2}✓ correct
  3. C. 23\dfrac{2}{3}
  4. D. 16\dfrac{1}{6}

Even outcomes on a die: {2,4,6}\{2, 4, 6\} — that is 33 outcomes.

Total outcomes: 66.

Probability =36=12= \dfrac{3}{6} = \dfrac{1}{2}.

Problem #0948 US SAT

Problem 5 — Events

If P(A)=0.3P(A) = 0.3, then P(Aˉ)P(\bar{A}) (the probability that AA does not occur) is:

Show answer & worked solution
  1. A. 0.30.3
  2. B. 11
  3. C. 0.70.7✓ correct
  4. D. 00

P(Aˉ)=1−P(A)=1−0.3=0.7P(\bar{A}) = 1 - P(A) = 1 - 0.3 = 0.7.

Problem #0826 International

Problem 6 — Probability

A bag contains 2020 coloured counters. The table shows how many counters there are of each colour. ColourRedBlueGreenYellowNumber5384\begin{array}{l|cccc} \text{Colour} & \text{Red} & \text{Blue} & \text{Green} & \text{Yellow} \\ \hline \text{Number} & 5 & 3 & 8 & 4 \end{array} One counter is taken at random. What is the probability that it is green?

Show answer & worked solution
  1. A. 25\frac{2}{5}✓ correct
  2. B. 35\frac{3}{5}
  3. C. 23\frac{2}{3}
  4. D. 14\frac{1}{4}

There are 88 green counters out of 2020 in total: P(green)=820P(\text{green}) = \frac{8}{20} Simplify the fraction by dividing top and bottom by 44: 820=25\frac{8}{20} = \frac{2}{5}

Problem #0927 US SAT

Problem 7 — Probability

The table shows the 5050 members of a school club, grouped by grade level and by whether they play a musical instrument.

PlaysDoesn’tTotalGrade 981220Grade 10121830Total203050\begin{array}{l|c|c|c} & \text{Plays} & \text{Doesn't} & \text{Total} \\ \hline \text{Grade 9} & 8 & 12 & 20 \\ \text{Grade 10} & 12 & 18 & 30 \\ \text{Total} & 20 & 30 & 50 \end{array}

If one member is selected at random, what is the probability that the member plays a musical instrument?

Show answer & worked solution
  1. A. 25\frac{2}{5}✓ correct
  2. B. 35\frac{3}{5}
  3. C. 23\frac{2}{3}
  4. D. 625\frac{6}{25}

A random selection makes every member equally likely, so the probability is the favorable count over the total count.

P(plays)=2050=25P(\text{plays}) = \frac{20}{50} = \frac{2}{5}

Problem #0764 FR Spé

Problem 8 — Probability

A workshop produces parts on two machines. Machine AA makes 60%60\% of the parts and machine BB makes the remaining 40%40\%. Among the parts from AA, 5%5\% are defective; among the parts from BB, 10%10\% are defective. A part is drawn at random from the day's production. What is the probability that it is defective?

Show answer & worked solution
  1. A. 0.070.07✓ correct
  2. B. 0.0750.075
  3. C. 0.150.15
  4. D. 0.040.04

Let DD be the event "the part is defective". By the law of total probability over the branches AA and BB: P(D)=P(A) PA(D)+P(B) PB(D)P(D) = P(A)\,P_A(D) + P(B)\,P_B(D) P(D)=0.60×0.05+0.40×0.10P(D) = 0.60 \times 0.05 + 0.40 \times 0.10 P(D)=0.03+0.04=0.07P(D) = 0.03 + 0.04 = 0.07 So the probability that the part is defective is 0.070.07.

Problem #0770 IB AA

Problem 9 — Probability

A factory uses two machines. Machine AA makes 60%60\% of the items and Machine BB makes the remaining 40%40\%. Of the items from Machine AA, 5%5\% are defective; of those from Machine BB, 10%10\% are defective. An item is selected at random and found to be defective. Find the probability that it was made by Machine AA.

Show answer & worked solution
  1. A. 37\dfrac{3}{7}✓ correct
  2. B. 47\dfrac{4}{7}
  3. C. 35\dfrac{3}{5}
  4. D. 3100\dfrac{3}{100}

Multiply along each branch to get the joint probabilities: P(A∩D)=0.6×0.05=0.03P(A\cap D)=0.6\times 0.05=0.03 P(B∩D)=0.4×0.10=0.04P(B\cap D)=0.4\times 0.10=0.04 The total probability of a defective item is P(D)=0.03+0.04=0.07P(D)=0.03+0.04=0.07 Apply the conditional probability formula: P(A∣D)=0.030.07=37P(A\mid D)=\frac{0.03}{0.07}=\frac{3}{7}

Problem #0468 International

Problem 10 — Probability

A coin is tossed 33 times. The probability of getting at least one head is:

Show answer & worked solution
  1. A. 18\dfrac{1}{8}
  2. B. 38\dfrac{3}{8}
  3. C. 12\dfrac{1}{2}
  4. D. 78\dfrac{7}{8}✓ correct

P(TTT)= ⁣(12)3=18P(\text{TTT}) = \!\left(\dfrac{1}{2}\right)^3 = \dfrac{1}{8}, so P(≥1H)=1−18=78P(\ge 1 H) = 1 - \dfrac{1}{8} = \dfrac{7}{8}.

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