Solve the problem.
-The following list of digits was taken from a table of random numbers. We will let them represent 50 tosses of 5 fair coins. If a digit is 0 - 4, it will be an H, and the digit 5 - 9 will be a T. Use this Simulation to estimate the probability of getting two heads on a toss of five coins. $\begin{array}{lllll}
18728 & 94741 & 32063 & 04963 & 54859 \
15592 & 71357 & 28152 & 11732 & 17173 \
89555 & 75520 & 08882 & 95295 & 72895 \
38894 & 58070 & 77371 & 10194 & 94917 \
64336 & 95103 & 90740 & 03577 & 62599
\end{array}$

$\begin{array}{lllll}
52052 & 24004 & 03845 & 11507 & 45233 \
27510 & 33761 & 86563 & 61729 & 39087 \
48061 & 59412 & 79969 & 11339 & 27067 \
27324 & 72723 & 22406 & 86253 & 26409 \
29970 & 95877 & 70975 & 99129 & 16866
\end{array}$

A) $\frac { 1 } { 4 }$
B) $\frac { 6 } { 25 }$
C) $\frac { 1 } { 2 }$
D) $\frac { 8 } { 25 }$

Question

Solve the problem.
-The following list of digits was taken from a table of random numbers. We will let them represent 50 tosses of 5 fair coins. If a digit is 0 - 4, it will be an H, and the digit 5 - 9 will be a T. Use this Simulation to estimate the probability of getting two heads on a toss of five coins. $\begin{array}{lllll}
18728 & 94741 & 32063 & 04963 & 54859 \
15592 & 71357 & 28152 & 11732 & 17173 \
89555 & 75520 & 08882 & 95295 & 72895 \
38894 & 58070 & 77371 & 10194 & 94917 \
64336 & 95103 & 90740 & 03577 & 62599
\end{array}$

$\begin{array}{lllll}
52052 & 24004 & 03845 & 11507 & 45233 \
27510 & 33761 & 86563 & 61729 & 39087 \
48061 & 59412 & 79969 & 11339 & 27067 \
27324 & 72723 & 22406 & 86253 & 26409 \
29970 & 95877 & 70975 & 99129 & 16866
\end{array}$

A) $\frac { 1 } { 4 }$
B) $\frac { 6 } { 25 }$
C) $\frac { 1 } { 2 }$
D) $\frac { 8 } { 25 }$

Quizplus · Accepted Answer

To estimate the probability of getting exactly two heads (H) in a toss of five coins using the given simulation, we convert each digit to H or T based on the rule: 0-4 is H, and 5-9 is T. Then, we count the number of times exactly two Hs appear in each group of five digits (or a "toss").After converting and counting, we find that there are 20 instances where exactly two Hs appear out of the total 50 tosses. Therefore, the estimated probability is $\frac{20}{50} = \frac{2}{5} = \frac{8}{20}$, which simplifies to $\frac{8}{25}$.

Solve the Problem A) 14\frac { 1 } { 4 }41​

Solve the Problem A) $\frac { 1 } { 4 }$