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GMAT Probability — practice questions

Explore 5 free Probability questions in GMAT Advanced Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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Probability questions and exercises with answers

Showing questions 1–5 of 5.

On average 15% of tomatoes are defective. In a batch of 20, the binomial probability of exactly 3 defective tomatoes is:Read the answer and worked solution →With 15% of tomatoes defective in batches of 20, the Poisson parameter $\lambda$ and $P(X=0)$ are:Read the answer and worked solution →For the binomial distribution table of defective tomatoes taken over $x=0$ to $5$, the mean and standard deviation are:Read the answer and worked solution →In a contest of 100 students, marks are normal with mean 64 and standard deviation 16. How many scored between 30% and 70% inclusive?Read the answer and worked solution →Binomial or Poisson?Read the answer and worked solution →

Worked practice questions

On average 15% of tomatoes are defective. In a batch of 20, the binomial probability of exactly 3 defective tomatoes is:

  1. 0.2428
  2. 0.0415
  3. 0.2240
  4. 0.1762
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Correct answer: A. 0.2428

1. Use $P(X=x)=\binom{n}{x}p^{x}q^{n-x}$ with $n=20$, $p=0.15$, $q=0.85$. 2. $\binom{20}{3}=1140$. 3. $P(X=3)=1140\times(0.15)^3\times(0.85)^{17}$. 4. This evaluates to $0.2428$. 5. $0.0415$ comes from substituting $n=6$ instead of $n=20$; $0.2240$ is the Poisson value for the same count. _Source: NECTA ACSEE 2023 Advanced Mathematics 142/2, Question 1: Probability_

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With 15% of tomatoes defective in batches of 20, the Poisson parameter $\lambda$ and $P(X=0)$ are:

  1. $\lambda=3$, $P=0.0498$
  2. $\lambda=0.15$, $P=0.8607$
  3. $\lambda=3$, $P=0.0388$
  4. $\lambda=20$, $P=0.0498$
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Correct answer: A. $\lambda=3$, $P=0.0498$

1. The Poisson approximation to a binomial uses $\lambda=np$. 2. Here $\lambda=20\times 0.15=3$. 3. Then $P(x)=\dfrac{\lambda^{x}e^{-\lambda}}{x!}$, so $P(0)=\dfrac{3^{0}e^{-3}}{0!}=e^{-3}$. 4. $e^{-3}=0.0498$ to four decimal places. 5. $0.0388$ is the binomial value for $x=0$, not the Poisson one; $\lambda=0.15$ mistakes the proportion for the mean count. _Source: NECTA ACSEE 2023 Advanced Mathematics 142/2, Question 1: Probability_

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For the binomial distribution table of defective tomatoes taken over $x=0$ to $5$, the mean and standard deviation are:

  1. 2.5662 and 1.4620
  2. 2.4454 and 1.5131
  3. 2.5662 and 1.5969
  4. 3.0000 and 1.5969
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Correct answer: A. 2.5662 and 1.4620

1. The table is truncated at $x=5$, so the mean must be computed from it directly, not from $np$. 2. Apply $E(X)=\sum x_i p(x_i)$ across $x=0$ to $5$, giving $2.5662$. 3. For the spread use $\delta=\sqrt{E(X^2)-[E(X)]^2}$. 4. That evaluates to $1.4620$. 5. Using $E(X)=np=3$ and $\delta=npq=1.5969$ ignores the truncation and is the common error; $2.4454$ and $1.5131$ belong to the Poisson table. _Source: NECTA ACSEE 2023 Advanced Mathematics 142/2, Question 1: Probability_

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