Math, asked by smokeinjone, 9 months ago

three balls are drawn from a bag containing 2 red and 5 black balls, if the random variable X represents the number of red balls drawn, then X can take values​

Answers

Answered by Anonymous
4

Answer:NSWER

Uru has 5 red and 2 black balls

No. of ways it can be drawn ={RR,RB,BR,BB}

n(S)=4

LEt X represent black balls

Possible values of X:X(RR)=0;X(RB)=1;X(BR)=1;X(BB)=2

Possible values of X are 0,1,2

P(X=0=  

4

1

​  

,P(X=2)=  

4

2

​  

,P(X=2)=  

4

1

​  

 

∑P  

i

​  

x  

i

​  

=P(X=0)+P(X=1)+P(X=2)=  

4

1

​  

+  

4

2

​  

+  

$

1

​  

=1

X is a random variable

Now, P(X=0=P(No black)

=  

7

​  

C  

2

​  

 

5

​  

C  

2

​  

 

​  

=  

42

20

​  

 

P(X=1)=P(1 black ball)=  

7

​  

C  

2

​  

 

5

​  

C  

1

​  

×  

2

​  

C  

1

​  

 

​  

=  

42

20

​  

 

P(X=2)=P(2 black balls)=  

7

​  

C  

2

​  

 

2

​  

C  

2

​  

 

​  

=  

42

2

​  

 

X=0;P(X=x)=  

42

20

​  

 

X=1;P(X=x)=  

42

20

Step-by-step explanation:

Answered by ruchibs1810
1

Answer:

The possible values on X that is red balls drawn is 0,1,2,3,4,5,6

Step-by-step explanation:

What is probability?

  • The area of mathematics known as probability deals with numerical representations of the likelihood that an event will occur or that a statement is true. An event's probability is a number between 0 and 1, where, roughly speaking, 0 denotes the event's impossibility and 1 denotes certainty.
  • There are four primary categories of probability: axiomatic, classical, empirical, and subjective.

Given that,

B bag has 2red and 5 black balls

No. of possible ways it can drawn is :{RRB,RBR,BRR,RBB,BRR,BRB,BBR,BBB}

n(S) = 8

Here X represent red balls

Possible values of X : X(RRB)=0;X(RBR) = 1;X(BRR) = 2;X(RBB) =3; X(BRR) = 4 ; X(BRB) = 5; X(BBR)=6

Therefore possible values of X are 0,1,2,3,4,5,6.

To learn more about probability refer to :

https://brainly.com/question/24756209

https://brainly.in/question/16017018

#SPJ2

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