Savita and Hamida are friends. What is the probability that both will have (i) different birthdays? (ii) the same birthday? (ignoring a leap year).
Step-by-Step Solution
Key Concept: Use the concept of equally likely outcomes and the multiplication principle. For two independent events (birthdays of two persons), the total number of possible ordered pairs of birthdays is $365 \times 365$. Count the favourable outcomes for each case and form the ratio.
1. Total possible outcomes\
Since each person can be born on any of the $365$ days (leap year ignored) and the birthdays are independent, the total number of ordered pairs of birthdays is\
$$\text{Total}=365 \times 365 = 133\,225.$$\
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2. (ii) Probability that they have the same birthday\
- For the birthdays to be the same, once Savita’s birthday is fixed (any of the $365$ days), Hamida must be born on that same day.\
- Hence the number of favourable outcomes = $365$ (one for each possible common day).\
- Probability\
$$P(\text{same birthday}) = \frac{\text{favourable}}{\text{total}} = \frac{365}{133\,225}=\frac{1}{365}.$$\
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3. (i) Probability that they have different birthdays\
- The complement of the event "same birthday" is "different birthdays".\
- Number of favourable outcomes for different birthdays = total outcomes $-$ same‑birthday outcomes\
$$\text{Favourable}=133\,225 - 365 = 132\,860.$$\
- Probability\
$$P(\text{different birthdays}) = \frac{132\,860}{133\,225}=\frac{364}{365}.$$\
- Alternatively, using conditional probability: after fixing Savita’s birthday, Hamida can be born on any of the remaining $364$ days, so\
$$P(\text{different}) = \frac{364}{365}.$$\
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4. Result\
$$P(\text{different birthdays}) = \frac{364}{365}, \qquad P(\text{same birthday}) = \frac{1}{365}.$$
Correct Answer: Different birthdays: $\frac{364}{365}$; Same birthday: $\frac{1}{365}$.