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Exercise 11.2 · Q3

Q.Find the probability mass function and the cumulative distribution function of the number of girl children in families with 44 children, assuming equal probabilities for boys and girls.

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With boys and girls equally likely, the number of girls XX among 44 children is Binomial(4,12)(4,\tfrac12); the pmf is (4x)(12)4\binom4x(\tfrac12)^4 and the cdf is its running sum.

Step 1. Identify the distribution. Each of the 44 children is independently a girl with probability 12\tfrac12, so X=X= number of girls ∼B ⁣(4,12)\sim B\!\left(4,\tfrac12\right).

Step 2. Write the pmf. f(x)=(4x)(12)4=(4x)16f(x)=\dbinom4x\left(\dfrac12\right)^4=\dfrac{\binom4x}{16}, x=0,1,2,3,4x=0,1,2,3,4.

f(0)=116, f(1)=416=14, f(2)=616=38, f(3)=416=14, f(4)=116f(0)=\dfrac1{16},\ f(1)=\dfrac4{16}=\dfrac14,\ f(2)=\dfrac6{16}=\dfrac38,\ f(3)=\dfrac4{16}=\dfrac14,\ f(4)=\dfrac1{16}.

Step 3. Check. 1+4+6+4+1=161+4+6+4+1=16, so the probabilities sum to 1616=1\dfrac{16}{16}=1 ✓.

Step 4. Build the cdf as the running sum.

F(0)=116F(0)=\dfrac1{16}; F(1)=116+416=516F(1)=\dfrac1{16}+\dfrac4{16}=\dfrac5{16}; F(2)=516+616=1116F(2)=\dfrac5{16}+\dfrac6{16}=\dfrac{11}{16}; F(3)=1116+416=1516F(3)=\dfrac{11}{16}+\dfrac4{16}=\dfrac{15}{16}; F(4)=1516+116=1F(4)=\dfrac{15}{16}+\dfrac1{16}=1.

✓Final answer

pmf: f(0)=116, f(1)=14, f(2)=38, f(3)=14, f(4)=116f(0)=\dfrac1{16},\ f(1)=\dfrac14,\ f(2)=\dfrac38,\ f(3)=\dfrac14,\ f(4)=\dfrac1{16}. cdf: F(0)=116, F(1)=516, F(2)=1116, F(3)=1516, F(4)=1F(0)=\dfrac1{16},\ F(1)=\dfrac5{16},\ F(2)=\dfrac{11}{16},\ F(3)=\dfrac{15}{16},\ F(4)=1.

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