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Parker's dog will be having four puppies. Parker performs a simulation by tossing a coin to model whether these puppies will be male or female.

- Let heads (H) = female puppy
- Let tails (T) = male puppy

The results of the simulation are:
[tex]\[
\begin{tabular}{|l|l|}
\hline
1. THHH & 6. THTT \\
\hline
2. HHHH & 7. HTTH \\
\hline
3. HTTT & 8. TITT \\
\hline
4. THHT & 9. HHTH \\
\hline
5. HHTH & 10. HHTT \\
\hline
\end{tabular}
\][/tex]

What is the estimated probability that at least two of the puppies will be female?

A. [tex]$\frac{6}{10}=60\%$[/tex]
B. [tex]$\frac{5}{10}=50\%$[/tex]
C. [tex]$\frac{4}{10}=40\%$[/tex]
D. [tex]$\frac{7}{10}=70\%$[/tex]

Sagot :

GinnyAnsweridn
To determine the estimated probability that at least two of the puppies will be female, let's analyze the simulation results given.

The results of the simulation are:
1. THHH
2. HHHH
3. HTTT
4. THHT
5. HHTH
6. THTT
7. HTTH
8. TITT
9. HHTH
10. HHTT

First, we need to count how many simulations have at least two heads (H), which represent female puppies.

1. THHH - 3 females
2. HHHH - 4 females
3. HTTT - 1 female
4. THHT - 2 females
5. HHTH - 3 females
6. THTT - 1 female
7. HTTH - 2 females
8. TITT - 1 invalid (contains 'I')
9. HHTH - 3 females
10. HHTT - 2 females

Counting the valid simulations with at least two female puppies:
1. THHH
2. HHHH
3. THHT
4. HHTH (5th)
5. HTTH (7th)
6. HHTH (9th)
7. HHTT (10th)

There are 7 such simulations.

Next, count the total number of valid simulations.

There are 10 simulations listed, but we need to exclude any invalid simulations. Here, the 8th simulation TITT contains 'I', which is not a valid result of a coin toss. After excluding this, we are left with 9 valid simulations:

So, we have 7 valid simulations with at least two female puppies out of 9 valid simulations.

Therefore, the estimated probability is:
[tex]\[ \frac{7}{10} = 70\% \][/tex]

Hence, the correct answer is:
[tex]\[ \boxed{7/10 = 70\%} \][/tex]

So, the correct option is:
D. [tex]\(\frac{7}{10}=70\%\)[/tex]
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