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Practice questions · Probability Distribution

Probability Distribution, Practice Questions

Six original Probability Distribution practice questions of rising difficulty, each with a complete worked solution. They cover a binomial probability P(X=r)P(X=r), the mean and standard deviation of a binomial, P(X1)P(X\ge 1) by the complement, a basic normal probability P(X>a)P(X>a), a normal probability between two values, and finding the mean of a normal distribution from a given probability.

Attempt each under timing, then check every line.

How to use these practice questions

The six questions below rise in difficulty across the whole chapter, from a single binomial probability to finding an unknown mean of a normal distribution. Give yourself roughly five to eight minutes per question and work on paper first, writing every line the way you would in the real exam, the distribution and its parameters, the formula you are using, a clear substitution, then the final probability.

Keep the standard normal distribution table beside you for the normal questions, and resist the urge to peek.

Only once you have committed to a full answer should you open the solution and mark yourself line by line. When your working differs from ours, stop at the exact line where the two part company; that single step is almost always where the mark was lost.

Because Add Math is marked analytically, a correctly quoted formula and a correct ZZ-value still earn credit even when the final rounding slips, so always write the formula before you substitute, and remember every probability must lie between 00 and 11.

Six practice questions

Q1[3 marks]

A discrete random variable XX follows a binomial distribution with n=5n=5 and p=0.2p=0.2, written XB(5,0.2)X\sim B(5,\,0.2). Find P(X=2)P(X=2).

Show worked solution

Use the binomial formula P(X=r)=nCrprqnrP(X=r)={}^{n}C_{r}\,p^{r}\,q^{\,n-r}, where q=1p=0.8q=1-p=0.8. Substitute n=5n=5, r=2r=2:

P(X=2)=5C2(0.2)2(0.8)3P(X=2)={}^{5}C_{2}\,(0.2)^{2}\,(0.8)^{3}
=10×0.04×0.512=0.2048=10\times 0.04\times 0.512=0.2048

Answer

P(X=2)=0.2048P(X=2)=0.2048. Check the pieces: 5C2=10{}^{5}C_{2}=10, (0.2)2=0.04(0.2)^{2}=0.04, (0.8)3=0.512(0.8)^{3}=0.512; their product is 0.20480.2048, which lies between 00 and 11 as a probability must.

Q2[3 marks]

A discrete random variable XX follows XB(25,0.2)X\sim B(25,\,0.2). Find the mean and the standard deviation of XX.

Show worked solution

For a binomial distribution the mean is μ=np\mu=np and the variance is σ2=npq\sigma^{2}=npq, with q=1p=0.8q=1-p=0.8. First the mean:

μ=np=25×0.2=5\mu=np=25\times 0.2=5

Now the variance, then take the square root for the standard deviation:

σ2=npq=25×0.2×0.8=4\sigma^{2}=npq=25\times 0.2\times 0.8=4
σ=4=2\sigma=\sqrt{4}=2

Answer

The mean is 55 and the standard deviation is 22. As a check, the standard deviation is smaller than the mean here, which is typical for a binomial with these parameters.

Q3[4 marks]

A discrete random variable XX follows XB(5,0.4)X\sim B(5,\,0.4). Find P(X1)P(X\ge 1).

Show worked solution

Adding P(X=1)P(X=1) up to P(X=5)P(X=5) is long, so use the complement: P(X1)=1P(X=0)P(X\ge 1)=1-P(X=0). With q=10.4=0.6q=1-0.4=0.6:

P(X=0)=5C0(0.4)0(0.6)5=(0.6)5=0.07776P(X=0)={}^{5}C_{0}\,(0.4)^{0}\,(0.6)^{5}=(0.6)^{5}=0.07776
P(X1)=10.07776=0.92224P(X\ge 1)=1-0.07776=0.92224

Answer

P(X1)=0.9222P(X\ge 1)=0.9222 (to four decimal places). The complement is the fast route whenever a question asks for “at least one”\text{``at least one''}: compute the single term P(X=0)P(X=0) and subtract from 11.

Q4[4 marks]

A continuous random variable XX is normally distributed with mean 5050 and standard deviation 1010, that is XN(50,102)X\sim N(50,\,10^{2}). Find P(X>65)P(X>65).

Show worked solution

Standardise with Z=XμσZ=\dfrac{X-\mu}{\sigma}, converting X=65X=65 to a ZZ-value using μ=50\mu=50 and σ=10\sigma=10:

Z=655010=1510=1.5Z=\frac{65-50}{10}=\frac{15}{10}=1.5
P(X>65)=P(Z>1.5)P(X>65)=P(Z>1.5)

From the standard normal distribution table, P(Z>1.5)=0.0668P(Z>1.5)=0.0668.

Answer

P(X>65)=0.0668P(X>65)=0.0668. Because 6565 lies above the mean, the tail probability is well below 0.50.5, which is the sensible order of size.

Q5[4 marks]

A continuous random variable XX is normally distributed with mean 6060 and standard deviation 88, that is XN(60,82)X\sim N(60,\,8^{2}). Find P(52<X<76)P(52<X<76).

Show worked solution

Standardise both endpoints with Z=XμσZ=\dfrac{X-\mu}{\sigma}, using μ=60\mu=60 and σ=8\sigma=8:

Z1=52608=1Z2=76608=2Z_{1}=\frac{52-60}{8}=-1 \qquad Z_{2}=\frac{76-60}{8}=2
P(52<X<76)=P(1<Z<2)P(52<X<76)=P(-1<Z<2)

From the table P(Z>2)=0.0228P(Z>2)=0.0228 and P(Z>1)=0.1587P(Z>1)=0.1587. Write the interval as P(Z<2)P(Z<1)P(Z<2)-P(Z<-1), and use the symmetry P(Z<1)=P(Z>1)P(Z<-1)=P(Z>1):

P(1<Z<2)=(10.0228)0.1587P(-1<Z<2)=\big(1-0.0228\big)-0.1587
=0.97720.1587=0.8185=0.9772-0.1587=0.8185

Answer

P(52<X<76)=0.8185P(52<X<76)=0.8185. The interval straddles the mean and covers a wide range, so a large probability is exactly what we expect.

Q6[5 marks]

A continuous random variable XX is normally distributed with mean μ\mu and standard deviation 55. Given that P(X>30)=0.1587P(X>30)=0.1587, (a) find μ\mu, and (b) hence find P(X<20)P(X<20).

Show worked solution

(a) First find the ZZ-value for which P(Z>z)=0.1587P(Z>z)=0.1587. From the standard normal table this is z=1z=1.

Now standardise X=30X=30 and set its ZZ-value equal to 11:

30μ5=1    30μ=5    μ=25\frac{30-\mu}{5}=1 \;\Rightarrow\; 30-\mu=5 \;\Rightarrow\; \mu=25

(b) With μ=25\mu=25 and σ=5\sigma=5, standardise X=20X=20, then use symmetry:

Z=20255=1Z=\frac{20-25}{5}=-1
P(X<20)=P(Z<1)=P(Z>1)=0.1587P(X<20)=P(Z<-1)=P(Z>1)=0.1587

Answer

μ=25\mu=25 and P(X<20)=0.1587P(X<20)=0.1587. This equals the given P(X>30)P(X>30), which makes sense: 2020 and 3030 are each 55 units, one standard deviation, from the mean 2525, one on each side, so their tail probabilities match.

How to mark yourself like an examiner

Add Math is marked analytically, which means marks are attached to steps, not only to the final number. When you check your own script, award yourself credit the way a marker would: look for the correct distribution, the right formula or ZZ-value, and a clean final probability.

  • Method mark: did you quote the correct formula, the binomial P(X=r)=nCrprqnrP(X=r)={}^{n}C_{r}\,p^{r}q^{\,n-r}, the mean npnp and variance npqnpq, or the standardisation Z=XμσZ=\dfrac{X-\mu}{\sigma}?
  • Parameter mark: for a binomial, are nn, pp and q=1pq=1-p correct, with the right power on each factor? For a normal, is σ\sigma (not σ2\sigma^{2}) used in the denominator?
  • For “at least one”\text{``at least one''}: did you use the complement 1P(X=0)1-P(X=0) rather than a long sum?
  • For a normal probability: is the ZZ-value correct, is the table read as a tail P(Z>z)P(Z>z), and is symmetry P(Z<a)=P(Z>a)P(Z<-a)=P(Z>a) applied for negative ZZ?
  • Answer mark: is the final probability between 00 and 11? Any value outside that range is an immediate signal that a step went wrong.

How a teacher helps

Marking yourself is powerful, but it is hard to see your own blind spots. In a one-to-one lesson our teacher watches the exact line where a mark slips away, a variance used where the standard deviation was needed, a table tail read from the wrong side, or an “at least”\text{``at least''} tackled the long way instead of by the complement, and corrects the habit on the spot.

Because our teachers are experienced, you work with someone who explains when to reach for the binomial and when for the normal. Lessons are taught in English, while SPM papers are set in both Malay and English, so we make sure the notation reads the same to you either way.

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Frequently asked questions

How do I decide between the binomial and the normal distribution?

Use the binomial when you have a fixed number of independent trials and you are counting how many succeed, a discrete whole-number count with XB(n,p)X\sim B(n,p). Use the normal when the variable is a continuous measurement described by a mean and a standard deviation, XN(μ,σ2)X\sim N(\mu,\sigma^{2}).

What is the fastest way to find P(X1)P(X\ge 1)?

Use the complement: P(X1)=1P(X=0)P(X\ge 1)=1-P(X=0). Instead of adding every case from 11 upwards, you compute the single term P(X=0)=qnP(X=0)=q^{\,n} and subtract it from 11.

This is the standard “at least one”\text{``at least one''} shortcut.

How do I read the table when the ZZ-value is negative?

The standard normal curve is symmetric about 00, so P(Z<a)=P(Z>a)P(Z<-a)=P(Z>a). Look up the positive value aa in the table and use it directly.

Always sketch the curve and shade the region you want before reading off a number.

What are the mean and variance of a binomial distribution?

For XB(n,p)X\sim B(n,p) the mean is μ=np\mu=np and the variance is σ2=npq\sigma^{2}=npq, where q=1pq=1-p. The standard deviation is σ=npq\sigma=\sqrt{npq}; take the square root only at the very end.

Source:SRC-DSKP-EN

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026

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