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Data Interview Question

Equal Coin Toss Outcomes

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Solution & Explanation

To determine the probability that two individuals, each flipping a coin four times, end up with the same number of heads, we need to consider the binomial distribution. Each coin flip is an independent event with two possible outcomes: heads or tails, each with a probability of 0.5.

1. Understanding the Binomial Distribution

For each individual flipping a coin four times, the number of heads follows a binomial distribution with parameters n=4n = 4 (number of trials) and p=0.5p = 0.5 (probability of getting heads in each trial). The probability mass function (PMF) for a binomial distribution is given by:

P(X=k)=(nk)pk(1p)nkP(X = k) = \binom{n}{k} p^k (1-p)^{n-k}

Where:

  • (nk)\binom{n}{k} is the binomial coefficient, representing the number of ways to choose kk successes (heads) out of nn trials.
  • pp is the probability of success (getting a head) in each trial.

2. Calculate Probabilities for Each Outcome

For n=4n = 4 and p=0.5p = 0.5, the probabilities for different numbers of heads kk (ranging from 0 to 4) are:

  • P(X=0)=(40)(0.5)0(0.5)4=116P(X = 0) = \binom{4}{0} (0.5)^0 (0.5)^4 = \frac{1}{16}
  • P(X=1)=(41)(0.5)1(0.5)3=416P(X = 1) = \binom{4}{1} (0.5)^1 (0.5)^3 = \frac{4}{16}
  • P(X=2)=(42)(0.5)2(0.5)2=616P(X = 2) = \binom{4}{2} (0.5)^2 (0.5)^2 = \frac{6}{16}
  • P(X=3)=(43)(0.5)3(0.5)1=416P(X = 3) = \binom{4}{3} (0.5)^3 (0.5)^1 = \frac{4}{16}
  • P(X=4)=(44)(0.5)4(0.5)0=116P(X = 4) = \binom{4}{4} (0.5)^4 (0.5)^0 = \frac{1}{16}

3. Calculate the Probability of Equal Outcomes

To find the probability that both individuals get the same number of heads, we sum the probabilities of both individuals having the same outcome for each possible number of heads:

P(same number of heads)=k=04P(X1=k)P(X2=k)P(\text{same number of heads}) = \sum_{k=0}^{4} P(X_1 = k) \cdot P(X_2 = k)

Substituting in the probabilities calculated:

\begin{align*} P(\text{same number of heads}) &= \left(\frac{1}{16}\right)^2 + \left(\frac{4}{16}\right)^2 + \left(\frac{6}{16}\right)^2 + \left(\frac{4}{16}\right)^2 + \left(\frac{1}{16}\right)^2 \\ &= \frac{1}{256} + \frac{16}{256} + \frac{36}{256} + \frac{16}{256} + \frac{1}{256} \\ &= \frac{70}{256} \\ &= \frac{35}{128} \\ &\approx 0.2734 \end{align*}

Thus, the probability that both individuals end up with the same number of heads after flipping a coin four times is approximately 0.2734, or 27.34%.