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A random variable X has the probability distribution:   

X\: :

1 2 3 4 5 6 7 8

P(X):

0.15 0.23 0.12 0.10 0.20 0.08 0.07

0.05

For the events E= \left \{ X is a prime number \left \ \right \} and F=\left \{ X< 4 \right \}, the probability P(E\cup F)  is:

  • Option 1)

    0.35

  • Option 2)

    0.77

  • Option 3)

    0.87

  • Option 4)

    0.50

 

Answers (1)

best_answer

 As we learnt in 

Probability of Distribution Of the Random Variable -

X :       0                 1                    2                      3        .......... r

P(X) :  P(0)           P(1)              P(2)                P(3)      ...........P(r)

such that 

 \dpi{100} \sum_{r =0}^{r }P(X)=1  

-

 

Set E = \left \{ 2, 3, 5, 7 \right \}

Set F = \left \{ 1, 2, 3 \right \}

Set \left ( E\cup F \right ) = \left \{ 1, 2, 3, 5, 7 \right \}

P \left ( E\cap F \right ) = P\left ( 1 \right )+\left P( 2 \right )+P\left ( 3 \right )+P\left ( 5 \right )+P\left ( 7 \right )

= 0.15+0.23+0.12+0.2+0.07

= 0.77


Option 1)

0.35

Incorrect

Option 2)

0.77

Correct

Option 3)

0.87

Incorrect

Option 4)

0.50

Incorrect

Posted by

divya.saini

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