Question

Three numbers are chosen from the first six natural numbers. Find the probability distribution of the
largest of the three numbers selected.​

Answer 1

Answer:

2 4 6 8 10 12 14 16 18 20 okyuu

Answer 2

The probability distribution of X for 3,4,5,6 is $\frac{1}{20}, \frac{3}{20}, \frac{6}{20}, \frac{10}{20}$

Step-by-step explanation:

we have first 6 natural numbers be 1,2,3,4,5,6

let  x be the largest of the three numbers  obtained

but x can take values 3,4,5,6

(1, 2 can not be selected since  among 3 numbers the greatest is required )

P(x) = P (3 and other numbers which are less than 3) i.e., (1,2,3)

= $\frac{1}{^6C_3} = \frac{1}{20}$

P (x=4)= P (4 and other numbers which are less than 4 ) i.e (1,2,4) (2,3,4) (1,3,4)

= $\frac{3}{^6C_3} = \frac{3}{20}$

P (x=5)= P (5 and other numbers which are less than 5 ) i.e  (1,2,5) (1,3,5) (1,4,5) (2,3,5) (2,4,5) (3,4,5)

$\frac{6}{^6C_3} = \frac{6}{20}$

P (x=6)= P (6 and other numbers which are less than 6 ) i.e  (1,2,6) (1,3,6) (1,4,6) (1,5,6) (2,3,6) (2,4,6) (2,5,6) (3,4,6) (3,5,6) (4,5,6)

$\frac{10}{^6C_3} = \frac{10}{20}$

hence The probability distribution of X for 3,4,5,6 is $\frac{1}{20}, \frac{3}{20}, \frac{6}{20}, \frac{10}{20}$

#Learn more:

Find the probability that three numbers chosen at random from first nine natural numbers from an AP?????

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