Solved

Using the Formula for the Multinomial Distribution, the Probability Of X

Question 72

Multiple Choice

Using the formula for the multinomial distribution, the probability of X1=2,X2=1,X3=2X _ { 1 } = 2 , X _ { 2 } = 1 , X _ { 3 } = 2 when n=5\mathrm { n } = 5 , p1=0.4,p2=0.5,p3=0.1\mathrm { p } _ { 1 } = 0.4 , \mathrm { p } _ { 2 } = 0.5 , \mathrm { p } _ { 3 } = 0.1 is


A) {5!/(2!×1!×2!) }2×0.42×0.51×0.12\{ 5 ! / ( 2 ! \times 1 ! \times 2 ! ) \} _ { 2 } \times 0.4 ^ { 2 } \times 0.5 ^ { 1 } \times 0.1 ^ { 2 }
B) 2!×1!×2!×0.42×0.51×0.122 ! \times 1 ! \times 2 ! \times 0.4 ^ { 2 } \times 0.5 ^ { 1 } \times 0.1 ^ { 2 }
C) 5!×0.42×0.51×0.125 ! \times 0.4 ^ { 2 } \times 0.5 ^ { 1 } \times 0.1 ^ { 2 }
D) {5!/(0.4!×0.5!×0.1!) }×20.4×10.5×20.1\{ 5 ! / ( 0.4 ! \times 0.5 ! \times 0.1 ! ) \} \times 2 ^ { 0.4 } \times 1 ^ { 0.5 } \times 2 ^ { 0.1 }

Correct Answer:

verifed

Verified

Unlock this answer now
Get Access to more Verified Answers free of charge

Related Questions