Probability - Inclusion and Exclusion

Question

For a group of 8 people, find the probability that all 4 seasons (winter, spring, summer, fall) occur at least once each among their birthdays, assuming that all seasons are equally likely.

Comments

Is it 0.623? :/

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Yes right answer :)

Answer 1

Answer: 0.623

Explanation:

Given that seasons are equally likely and all 4 seasons (winter, spring, summer, fall) occur at least once each among their birthdays

Total outcomes: Each person is allotted a season out of 4. Hence 4⁸ possibilities

Number of Outcomes that one or more season has no student having their birthday:

Using Inclusion and Exclusion, ⁴C₁*3⁸ - ⁴C₂*2⁸ + ⁴C₃*1⁸  [i.e. Exclude 1 season - Exclude 2 season + exclude 3 season, also note that we can't exclude all the 4 seasons]

Probability that one or more season has no student having their birthday:

(⁴C₁*3⁸ - ⁴C₂*2⁸ + ⁴C₃*1⁸  ) / 4⁸ = 0.377

Required probability that all 4 seasons (winter, spring, summer, fall) occur at least once each among their birthdays:

1-0.377=0.623

Answer 2

$Seasons = \left \{ Winter , \ spring, \ summer, \ fall \right \}$

$Number \  of \ people = 8$

$X$ be a variable such that it denotes the number of seasons which consists of birthdays of all 8 people,

All seasons are equally likely,

$P(Winter) = P(spring) = P(Summer) = P(Fall) = $$ \large \frac{1}{4}$

$P\left \{ X = i \right \}$ is probability of birthdays occurring in $i$ seasons,

$P\left \{ X = 1 \right \} =$$\Large \binom{4}{1} \left ( \frac{1}{4} \right )^8 = 0.00006103515$

$P\left \{ X = 2 \right \} =$$\Large \binom{4}{2} \left ( \frac{1}{2} \right )^8 = 0.02345$

$P\left \{ X = 3 \right \} =$$\Large \binom{4}{3} \left ( \frac{3}{4} \right )^8 = 0.4004$

$\large  P\left \{ X = 4 \right \} =$$\large 1 - \left ( P\left \{ X = 3 \right \} - P\left \{ X = 2 \right \} + P\left \{ X = 1 \right \}   \right ) $

$\large P\left \{ X = 4 \right \} =$$\large 1 - \left ( 0.4004 - 0.02345 +  0.0000610  \right ) $

$\large P\left \{ X = 4 \right \} = 1 - 0.377011 = 0.623$