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2a. A cat has four kittens. Draw a tree diagram to show all the possible combinations of males and females. (Assume [tex]P(\text{male}) = P(\text{female})[/tex])

2b. Calculate the probability of getting:
i. all female,
ii. two females and two males,
iii. at least one female,
iv. more females than males.



Answer :

GinnyAnswer
### Part 2a: Tree Diagram

To represent the possible combinations of having male and female kittens, we can draw a tree diagram. Each level of the tree will show the gender of one kitten. There are four levels in the tree because there are four kittens.

Here’s a step-by-step breakdown of the tree diagram:

1. First Kitten:
- M (Male)
- F (Female)

2. Second Kitten (for each possibility of the first kitten):
- If the first kitten is M:
- MM (Male, Male)
- MF (Male, Female)
- If the first kitten is F:
- FM (Female, Male)
- FF (Female, Female)

3. Third Kitten (for each possibility of the first and second kittens):
- If the second combination is MM:
- MMM (Male, Male, Male)
- MMF (Male, Male, Female)
- If the second combination is MF:
- MFM (Male, Female, Male)
- MFF (Male, Female, Female)
- If the second combination is FM:
- FMM (Female, Male, Male)
- FMF (Female, Male, Female)
- If the second combination is FF:
- FFM (Female, Female, Male)
- FFF (Female, Female, Female)

4. Fourth Kitten (for each combination of the first, second, and third kittens):
- If the third combination is MMM:
- MMMM (Male, Male, Male, Male)
- MMMF (Male, Male, Male, Female)
- If the third combination is MMF:
- MMFM (Male, Male, Female, Male)
- MMFF (Male, Male, Female, Female)
- ... and so on for each possible combination up to FF.

This results in a total of 2^4 = 16 combinations. Here they all are:

- MMMM
- MMMF
- MMFM
- MMFF
- MFMM
- MFMF
- MFFM
- MFFF
- FMMM
- FMMF
- FMFM
- FMFF
- FFMM
- FFMF
- FFFM
- FFFF

### Part 2b: Calculating Probabilities

To find the probability of each requirement, we use the total number of possible combinations, which is 16.

#### i. Probability of Getting All Female

- There is only one combination: FFFF.
- Total favorable outcomes = 1
- Total possible outcomes = 16

[tex]\[ \text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total possible outcomes}} = \frac{1}{16} = 0.0625 \][/tex]

#### ii. Probability of Having Two Females and Two Males

- The combinations are MMFF, MFMF, MFFM, FMFF, FFMF, FMMF.
- Total favorable outcomes = 6
- Total possible outcomes = 16

[tex]\[ \text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total possible outcomes}} = \frac{6}{16} = 0.375 \][/tex]

#### iii. Probability of Having At Least One Female

- The combinations that do not have at least one female (hence all male) are: MMMM.
- Exclude the ‘all male’ combination leaves 15 out of 16 which have at least one female.
- Total favorable outcomes = 15
- Total possible outcomes = 16

[tex]\[ \text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total possible outcomes}} = \frac{15}{16} = 0.9375 \][/tex]

#### iv. Probability of Having More Females Than Males

- The combinations where the count of females is greater than the count of males (3 females, 1 male or 4 females) are: FFFF, FFFM, FFMF, FMFF, MFFF.
- Total favorable outcomes = 5
- Total possible outcomes = 16

[tex]\[ \text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total possible outcomes}} = \frac{5}{16} = 0.3125 \][/tex]

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