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Question 16 of 43

Q.Find the constant KK for the following discrete probability distribution. Hence obtain mean of this distribution: P(x)=K⋅4Px,x=0,1,2,3,4P(x) = K \cdot {}^{4}P_{x}, \quad x = 0, 1, 2, 3, 4

Gujarat GsebGujarat Board (GSEB) HSC Commerce Board 2020Subjective· 3mImportance★★★★★
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K∑4Px=1⇒K=1/65K\sum {}^4P_x = 1 \Rightarrow K = 1/65; mean =165∑x 4Px=19665=3.02= \frac{1}{65}\sum x\,{}^4P_x = \frac{196}{65} = 3.02.

Here P(x)=K⋅4PxP(x) = K\cdot {}^4P_x for x=0,1,2,3,4x = 0,1,2,3,4, where 4Px=4!(4−x)!{}^4P_x = \dfrac{4!}{(4-x)!}:

4P0=1, 4P1=4, 4P2=12, 4P3=24, 4P4=24{}^4P_0 = 1,\ {}^4P_1 = 4,\ {}^4P_2 = 12,\ {}^4P_3 = 24,\ {}^4P_4 = 24

Step 1 — find KK. Total probability is 1:

∑x=04P(x)=K∑4Px=K(1+4+12+24+24)=65K=1  ⟹  K=165\sum_{x=0}^{4} P(x) = K\sum {}^4P_x = K(1+4+12+24+24) = 65K = 1 \implies K = \frac{1}{65}

Step 2 — probability distribution and mean.

| xx | 0 | 1 | 2 | 3 | 4 |

|---|---|---|---|---|---| …

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