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Sheetal Beharry
Asked: 2022-11-11T16:54:14+05:30 In:

A die thrown 500 times and the outcomes are noted as given below :
image
If a die is thrown at random , find the probability of getting
(i) 1 (ii) 2 (iii) 3 (iv) 4 (v) 5 (vi) 6.

A die thrown 500 times and the outcomes are noted as given below :
image
If a die is thrown at random , find the probability of getting
(i) 1 (ii) 2 (iii) 3 (iv) 4 (v) 5 (vi) 6.
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  1. Total number of trials = 500.
    In a random throw of a die , let `E_(1) , E_(2) , E_(3) , E_(4) , E_(5)` and `E_(6)` be the events of getting 1 , 2 ,3 , 4 , 5 and 6 respectively . Then ,
    (i) P (getting 1 ) = `P(E_(1)) = (“number of times 1 appears”)/(“total number of trials”)`
    `= (95)/(500) = (19)/(100) = 0.19`
    (ii) P (getting 2) = `P (E_(2)) = (“number of times 2 appears”)/(“total number of trials”)`
    `= (80)/(500) = (16)/(100) = 0.16`
    (iii) P(getting 3) = `P (E_(3))= (“number of times 3 appears”)/(“total number of trials”)`
    `(84)/(500) = 0.168.`
    (iv) P (getting 4) = `P(E_(4)) = (“number of times 4 appears”)/(“total number of trials”)`
    `= (68)/(500) = 0.136`
    (v) P(getting 5) `= P(E_(5)) = (“number of times 5 appears”)/(“total number of trials”)`
    `= (70)/(500) = (7)/(50) = 0.14`
    (vi) P(getting 6) = `P(E_(6)) = (“number of times 6 appears”)/(“total number of trials”)`
    `= (103)/(500) = 0.206`.

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