Question

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.”
32766 53855 34591 27732
47406 31022 25144 72662
03087 35521 26658 81704
56212 72345 44019 65311

Answer 1

To use the random number table, Eva can group the digits into pairs (e.g. 32, 76, 66) and consider each pair as one trial of flipping the coin. If the pair represents “heads” (an odd number), she can count it as a success, and if it represents “tails” (an even number), she can count it as a failure. Then, she can use the binomial distribution formula:

P(X ≥ 3) = 1 - P(X < 3) = 1 - [P(X = 0) + P(X = 1) + P(X = 2)]

where X is the number of times getting “heads” in five flips, and P(X = k) is the probability of getting “heads” exactly k times out of five. The probability of one trial being a success is 1/2, and the probability of a failure is also 1/2. Therefore,

P(X = k) = (5 choose k) (1/2)^5

where (5 choose k) is the number of ways to choose k successes out of five trials, and it can be calculated as:

(5 choose k) = 5! / [k! (5 - k)!]

Using this formula, Eva can calculate the probabilities for each possible value of X:

P(X = 0) = (5 choose 0) (1/2)^5 = 1/32
P(X = 1) = (5 choose 1) (1/2)^5 = 5/32
P(X = 2) = (5 choose 2) (1/2)^5 = 10/32

Then, she can substitute these values into the formula for P(X ≥ 3):

P(X ≥ 3) = 1 - (1/32 + 5/32 + 10/32) = 16/32 = 1/2

Therefore, the probability of tossing “heads” at least three times out of five is 1/2 or 50%.