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Suppose a fair coin is tossed 10 times. What is the probability of obtaining exactly 4 heads in 10 tosses?

b. Suppose a fair die is rolled 10 times. What is the probability of getting "6" exactly 4 times?

User Buzkall
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Final answer:

The probability of getting exactly 4 heads in 10 tosses of a fair coin is approximately 20.51%. The probability of getting '6' exactly 4 times in 10 rolls of a fair die is approximately 5.43%. This is calculated using the binomial probability formula.

Step-by-step explanation:

The subject of this question is mathematics, specifically the topic of probability, where the student is asked to calculate the probability of getting a certain number of heads when tossing coins, and getting a specific outcome when rolling dice a number of times.

These problems involve the use of the binomial probability formula and combinatorial principles.

Probability of Getting Exactly 4 Heads in 10 Tosses of a Fair Coin

For a fair coin tossed 10 times, the probability of obtaining exactly 4 heads is given by the binomial probability formula:

P(X=k) = C(n, k) × (p^k) × ((1-p)^(n-k))

Where:

C(n, k) is the number of combinations of n items taken k at a time.

p is the probability of getting a head in one toss.

k is the number of heads we want to get (which is 4 in this case).

n is the total number of tosses (which is 10).

For this fair coin p = 0.5, so the probability is:

P(4 heads in 10 tosses) = C(10, 4) × (0.5^4) × (0.5^(10-4)) = 210 × 0.0625 × 0.0625 ≈ 0.2051 or 20.51%

Probability of Getting '6' Exactly 4 Times in 10 Rolls of a Fair Die

Similarly, for a fair six-sided die rolled 10 times, the probability of rolling '6' exactly 4 times can be calculated by using the same formula, with p = 1/6 (since there is one '6' in a die of six faces), thus:

P(6 rolled 4 times in 10 rolls) = C(10, 4) × (1/6)^4 × (5/6)^(10-4) ≈ 0.0543 or 5.43%

User Rhona
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