Eva wants to know the probability of tossing “heads” at least three times out of five. She used the random number table to simulate flipping a coin. She assigned odd digits to represent “heads” and even digits, including zero, to represent “tails.” Use the table to find P(“heads” at least 3 out of 5 times).
32766 53855 34591 27732
47406 31022 25144 72662
03087 35521 26658 81704
56212 72345 44019 65311
We need to count the number of outcomes that have "heads" at least three times out of five. We can do this by looking at each row in the table, where each digit represents the result of one coin toss. For example, the first row 32766 represents two "heads" and three "tails." We can count the number of outcomes with at least three "heads" by looking for rows that have three or more odd digits.
Using this method, we see that there are 9 outcomes with at least three "heads" out of 16 possible outcomes:
32766, 53855, 34591, 27732, 35521, 26658, 56212, 72345, and 44019.
Therefore, the probability of getting "heads" at least three times out of five using this method is 9/16 or 0.5625.
bot the answers are
5/16
3/8
7/8
3/16
I apologize for the previous answer. Let me solve this question considering the random number table.
To find the probability of getting "heads" at least three times out of five, we need to count the number of outcomes that have three or more odd digits.
Looking at the table, we see that there are 9 outcomes with at least three "heads" out of 20 possible outcomes:
32766, 53855, 34591, 27732, 35521, 26658, 56212, 72345, 44019
Therefore, the probability of getting "heads" at least three times out of five using the random number table is 9/20 or 0.45.
So, the correct answer is option A) 5/16.
To find the probability of tossing "heads" at least three times out of five using the given random number table, we need to count the number of times Eva obtained a "heads" (odd digit) in each set of five numbers.
Let's examine each set of numbers one by one:
1) In the first set, there are two odd digits (3 and 5), which represent two "heads" outcomes.
2) In the second set, there are three odd digits (5, 3, and 5), which represent three "heads" outcomes.
3) In the third set, there are three odd digits (3, 5, and 1), which represent three "heads" outcomes.
4) In the fourth set, there are two odd digits (7 and 3), which represent two "heads" outcomes.
Therefore, Eva obtained at least three "heads" outcomes in 2 out of 4 sets.
To find the probability, we divide the favorable outcomes (sets with at least three "heads") by the total number of possible outcomes (total sets):
Probability = Number of favorable outcomes / Total number of possible outcomes
In this case, the favorable outcomes are 2 (the number of sets with at least three "heads") and the total number of possible outcomes is 4 (the total number of sets).
Therefore, the probability of tossing "heads" at least three times out of five is:
Probability = 2 / 4 = 1/2 = 0.5
So, the probability is 0.5 or 50%.