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Question 105 of 105
Q.

(a) A random variable X has the following probability mass function.

x123456
f(x)k2k6k5k6k10k

Find (i) P(2<X<6)P(2<X<6) (ii) P(2≤X<5)P(2\le X<5) (iii) P(X≤4)P(X\le4) (iv) P(3<X)P(3<X)

OR

(b) At a water fountain, water attains a maximum height of 4 m at horizontal distance of 0.5 m from its origin. If the path of water is a parabola, find the height of water at a horizontal distance of 0.75 m from the point of origin.

Puducherry TnboardTamil Nadu HSC (DGE) Board 2026Subjective· 5mImportance★★★★★
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(a) Finds kk from the pmf's total-probability condition, then sums the pmf over the required ranges for each of the four probabilities; (b) fits a downward parabola with vertex at the fountain's peak through the origin, then evaluates it at the required horizontal distance. Both alternatives answered below.

(a) pmf of XX and the four probabilities

1. Find kk. Total probability must equal 11:

k+2k+6k+5k+6k+10k=30k=1 ⟹ k=130k+2k+6k+5k+6k+10k=30k=1\ \Longrightarrow\ k=\dfrac1{30}

2. Individual probabilities.

xx123456
f(x)f(x)130\frac1{30}230\frac2{30}630\frac6{30}530\frac5{30}630\frac6{30}1030\frac{10}{30}

Check: 1+2+6+5+6+1030=3030=1\dfrac{1+2+6+5+6+10}{30}=\dfrac{30}{30}=1✓.

3. (i) P(2<X<6)=P(X=3,4,5)P(2<X<6)=P(X=3,4,5).

f(3)+f(4)+f(5)=6+5+630=1730f(3)+f(4)+f(5)=\dfrac{6+5+6}{30}=\dfrac{17}{30}

4. (ii) P(2≤X<5)=P(X=2,3,4)P(2\le X<5)=P(X=2,3,4).

f(2)+f(3)+f(4)=2+6+530=1330f(2)+f(3)+f(4)=\dfrac{2+6+5}{30}=\dfrac{13}{30}

5. (iii) P(X≤4)=P(X=1,2,3,4)P(X\le4)=P(X=1,2,3,4).

f(1)+f(2)+f(3)+f(4)=1+2+6+530=1430=715f(1)+f(2)+f(3)+f(4)=\dfrac{1+2+6+5}{30}=\dfrac{14}{30}=\dfrac7{15}

6. (iv) P(3<X)=P(X=4,5,6)P(3<X)=P(X=4,5,6).

f(4)+f(5)+f(6)=5+6+1030=2130=710f(4)+f(5)+f(6)=\dfrac{5+6+10}{30}=\dfrac{21}{30}=\dfrac7{10}

(b) Height of the fountain's water at x=0.75x=0.75 m

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