Chapter 7.3, Problem 7.77E

(a)

To determine

To Calculate: the probability that they have different birthdays and the probability that they have the same birthday.

Answer to Problem 7.77E

  $\begin{array}{c}\text{P}\left(\text{Different}\text{birthday}\right)\text{=0}\text{.9973}\\ \text{P}\left(\text{same}\text{birthday}\right)\text{=0}\text{.0027}\end{array}$

Explanation of Solution

Formula used:

  $\text{Probability}\text{=}\frac{\text{Favourable Cases}}{\text{Total}\text{Cases}}$

Compliment Rule

  $\text{P}\left({\text{A}}^c\right)=1-\text{P}\left(\text{A}\right)$

Calculation:

Suppose there is no leap year and therefore a year have 365 days

There are 365 days to select from for the first person and 365 days for the second person

  $365\times 364=133225$

If the two people have different birthdays, then there are 365 days to select from for the first person and 364 days for the second person

  $365\times 364=132860$

132860 of the 133225 possible pairs of birthday will then result in the two people having different birthdays.

  $\begin{array}{c}\left(\text{Different}\text{birthday}\right)\text{=}\frac{\text{365×364}}{\text{365×365}}\\ \text{=}\frac{\text{364}}{\text{365}}\\ \text{=0}\text{.9973}\end{array}$

  $\begin{array}{c}\text{P}\left(\text{same}\text{birthday}\right)\text{=1-P}\left(\text{different}\text{birthday}\right)\\ \text{=1-}\frac{\text{364}}{\text{365}}\\ \text{=0}\text{.0027}\end{array}$

(b)

To determine

To Calculate: the probability that all 3 have different birthdays and the probability that at least 2 of them have the same birthday.

Answer to Problem 7.77E

  $\begin{array}{c}\text{P}\left(\text{Different}\text{birthday}\right)\text{=0}\text{.9918}\\ \text{P}\left(\text{same}\text{birthday}\right)\text{=0}\text{.0082}\end{array}$

Explanation of Solution

Formula used:

  $\text{Probability}\text{=}\frac{\text{Favourable Cases}}{\text{Total}\text{Cases}}$

Compliment Rule

  $\text{P}\left({\text{A}}^c\right)=1-\text{P}\left(\text{A}\right)$

Calculation:

Suppose there is no leap year and therefore a year have 365 days

There are 365 days to select from for the first person and 365 days for the second person and 365 days for the third person

  $365\times 365\times 365={365}^3$

If the three people have different birthdays, then there are 365 days to select from for the first person and 364 days for the second person

  $365\times 364\times 363=48228180$

48228180 of the ${365}^3$ possible pairs of birthday will then result in the two people having different birthdays.

  $\begin{array}{c}\left(\text{Different}\text{birthday}\right)\text{=}\frac{\text{365×364}\times \text{363}}{{\text{365}}^3}\\ \text{=0}\text{.9918}\end{array}$

  $\begin{array}{c}\text{P}\left(\text{same}\text{birthday}\right)\text{=1-P}\left(\text{different}\text{birthday}\right)\\ \text{=1-}\frac{\text{365×364}\times \text{363}}{{\text{365}}^3}\\ \text{=1-0}\text{.9918}\\ \text{=0}\text{.0082}\end{array}$

(c)

To determine

To Calculate: the probability that all 30 have different birthday and the probability that at least 2 of them have the same birthday.

Answer to Problem 7.77E

  $\begin{array}{c}\text{P}\left(\text{Different}\text{birthday}\right)\text{=0}\text{.2937}\\ \text{P}\left(\text{same}\text{birthday}\right)\text{=0}\text{.7063}\end{array}$

Explanation of Solution

Formula used:

  $\text{Probability}\text{=}\frac{\text{Favourable Cases}}{\text{Total}\text{Cases}}$

Compliment Rule

  $\text{P}\left({\text{A}}^c\right)=1-\text{P}\left(\text{A}\right)$

Calculation:

Suppose there is no leap year and therefore a year have 365 days

There are 365 days to select from for the first person and 365 days for the second person and 365 days for the third person and so on.

  $365\times 365\times \mathrm{....}\times 356\times 365={365}^{30}$

If the three people have different birthdays, then there are 365 days to select from for the first person and 364 days for the second person

  $365\times 364\times 363\times 362\times \mathrm{....}\times 336$

$365\times 364\times 363\times 362\times \mathrm{....}\times 336$ Of the ${365}^{30}$ possible pairs of birthday will then result in the two people having different birthdays.

  $\begin{array}{c}\left(\text{Different}\text{birthday}\right)\text{=}\frac{365\times 364\times 363\times 362\times \mathrm{....}\times 336}{{365}^{30}}\\ \text{=0}\text{.2937}\end{array}$

  $\begin{array}{c}\text{P}\left(\text{same}\text{birthday}\right)\text{=1-P}\left(\text{different}\text{birthday}\right)\\ \text{=1-}\frac{365\times 364\times 363\times 362\times \mathrm{....}\times 336}{{365}^{30}}\\ \text{=1-0}\text{.2937}\\ \text{=0}\text{.7063}\end{array}$