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Sepasang suami istri merencanakan untuk mempunyai 4 orang anak. Jika variabel acak X menyatakan banyak perempuan, nilai dari P(X <= 2) adalah .....

Sepasang suami istri merencanakan untuk mempunyai 44 orang anak. Jika variabel acak X menyatakan banyak perempuan, nilai dari P(X≤2) P(X \leq 2) adalah .....

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Q. Sepasang suami istri merencanakan untuk mempunyai 44 orang anak. Jika variabel acak X menyatakan banyak perempuan, nilai dari P(X≤2) P(X \leq 2) adalah .....
  1. Define random variable and distribution: Define the random variable and the distribution.\newlineThe random variable XX represents the number of daughters in a family of four children. Assuming that the probability of having a daughter is the same as having a son, and each birth is independent of the others, XX follows a binomial distribution with parameters n=4n = 4 (number of trials) and p=0.5p = 0.5 (probability of success, i.e., having a daughter).
  2. Calculate probabilities of daughters: Calculate the probability of having 00, 11, or 22 daughters.\newlineWe need to calculate P(X=0)P(X = 0), P(X=1)P(X = 1), and P(X=2)P(X = 2) and then sum these probabilities to find P(X≤2)P(X \leq 2).\newlineThe probability mass function (PMF) for a binomial distribution is given by:\newlineP(X=k)=(nk)⋅pk⋅(1−p)n−kP(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k}\newlinewhere “n choose k”\text{“n choose k”} is the binomial coefficient calculated as n!k!⋅(n−k)!\frac{n!}{k! \cdot (n-k)!}.
  3. Calculate P(X=0)P(X = 0): Calculate P(X=0)P(X = 0). Using the PMF: P(X=0)=(40)×(0.5)0×(0.5)4−0P(X = 0) = \binom{4}{0} \times (0.5)^0 \times (0.5)^{4-0} P(X=0)=1×1×(0.5)4P(X = 0) = 1 \times 1 \times (0.5)^4 P(X=0)=0.0625P(X = 0) = 0.0625
  4. Calculate P(X=1)P(X = 1): Calculate P(X=1)P(X = 1). Using the PMF: P(X=1)=(41)×(0.5)1×(0.5)4−1P(X = 1) = \binom{4}{1} \times (0.5)^1 \times (0.5)^{4-1} P(X=1)=4×0.5×(0.5)3P(X = 1) = 4 \times 0.5 \times (0.5)^3 P(X=1)=4×0.5×0.125P(X = 1) = 4 \times 0.5 \times 0.125 P(X=1)=0.25P(X = 1) = 0.25
  5. Calculate P(X=2)P(X = 2): Calculate P(X=2)P(X = 2).\newlineUsing the PMF:\newlineP(X=2)=(42)×(0.5)2×(0.5)4−2P(X = 2) = \binom{4}{2} \times (0.5)^2 \times (0.5)^{4-2}\newlineP(X=2)=6×0.25×0.25P(X = 2) = 6 \times 0.25 \times 0.25\newlineP(X=2)=6×0.0625P(X = 2) = 6 \times 0.0625\newlineP(X=2)=0.375P(X = 2) = 0.375
  6. Sum probabilities for P(X≤2)P(X \leq 2): Sum the probabilities to find P(X≤2)P(X \leq 2).
    P(X≤2)=P(X=0)+P(X=1)+P(X=2)P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)
    P(X≤2)=0.0625+0.25+0.375P(X \leq 2) = 0.0625 + 0.25 + 0.375
    P(X≤2)=0.6875P(X \leq 2) = 0.6875

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