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Harpreet tosses two different coins simultaneously. What is the probability of getting at least one head?Read the answer and worked solution →Savita and Hamida are friends. Ignoring leap year, what is the probability that they have the same birthday?Read the answer and worked solution →One card is drawn from a well-shuffled deck of 52 cards. What is the probability that the card is not an ace?Read the answer and worked solution →Which of the following is an impossible event when a fair die is thrown once?Read the answer and worked solution →

Worked practice questions

Harpreet tosses two different coins simultaneously. What is the probability of getting at least one head?

  1. $\frac{1}{4}$
  2. $\frac{1}{2}$
  3. $\frac{3}{4}$
  4. $1$
Show answer and solution

Correct answer: C. $\frac{3}{4}$

The equally likely outcomes are $(H,H)$, $(H,T)$, $(T,H)$ and $(T,T)$. At least one head occurs in the first three outcomes, so there are 3 favourable outcomes out of 4. Thus the probability is $\frac{3}{4}$.

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Savita and Hamida are friends. Ignoring leap year, what is the probability that they have the same birthday?

  1. $\frac{1}{364}$
  2. $\frac{1}{365}$
  3. $\frac{364}{365}$
  4. $\frac{2}{365}$
Show answer and solution

Correct answer: B. $\frac{1}{365}$

Fix Savita's birthday on any day. Hamida can have any of 365 equally likely birthdays, and only 1 of these matches Savita's. **So the probability of the same birthday is $\frac{1}{365}$.**

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One card is drawn from a well-shuffled deck of 52 cards. What is the probability that the card is not an ace?

  1. $\frac{1}{13}$
  2. $\frac{4}{13}$
  3. $\frac{12}{13}$
  4. $\frac{48}{13}$
Show answer and solution

Correct answer: C. $\frac{12}{13}$

A deck has 4 aces, so non-aces are $52-4=48$. Hence the probability is $\frac{48}{52}=\frac{12}{13}$. Equivalently, it is $1-\frac{1}{13}=\frac{12}{13}$.

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