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Exercise 9.2 · Q2

Q.Give a real-life example of the following: Impossible and Sure Events.

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✓ Free question

An impossible event can never occur (P=0P=0); a sure event always occurs (P=1P=1) — shown with one dice-based example of each.

For a sample space SS of an experiment:

Impossible event: E=∅  ⇒  P(E)=n(E)n(S)=0n(S)=0\text{Impossible event: } E=\varnothing \;\Rightarrow\; P(E)=\dfrac{n(E)}{n(S)}=\dfrac{0}{n(S)}=0

Sure (certain) event: E=S  ⇒  P(E)=n(S)n(S)=1\text{Sure (certain) event: } E=S \;\Rightarrow\; P(E)=\dfrac{n(S)}{n(S)}=1

  1. Impossible event — real-life example. Roll a single fair die once; sample space S={1,2,3,4,5,6}S=\{1,2,3,4,5,6\}. Let E=E= "the die shows a number greater than 66". No outcome in SS satisfies this, so E=∅E=\varnothing, n(E)=0n(E)=0.

P(E)=n(E)n(S)=06=0P(E)=\frac{n(E)}{n(S)}=\frac{0}{6}=0

  1. Sure event — real-life example. Roll the same die once. Let F=F= "the die shows a number less than 77". Every outcome {1,2,3,4,5,6}\{1,2,3,4,5,6\} satisfies this, so F=SF=S, n(F)=6n(F)=6.

P(F)=n(F)n(S)=66=1P(F)=\frac{n(F)}{n(S)}=\frac{6}{6}=1

Self-check: E∪F=SE\cup F=S and E∩F=∅E\cap F=\varnothing; P(E)+P(F)=0+1=1P(E)+P(F)=0+1=1, consistent with FF being the complement of EE relative to a trivially-false condition.

✓Final answer

Impossible event: rolling a number >6>6 on a single fair die, P=0P=0.

Sure event: rolling a number <7<7 on a single fair die, P=1P=1.

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