Question

A die is thrown 1000 times with the frequencies for the outcomes 1,2,3,4,5
and 6 as given in the following table :



Find the probability of getting
0 an odd number
(ii) a prime number
(iii) a number greater than 4.
fiu) a multiple of 2

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Answer 1

i) P(getting odd no.)

= [P(1)+P(3)+P(5)]/1000

=(179+157+175)/1000

=511/1000

ii) P(getting prime no.)

=[P(2)+P(3)+P(5)]/1000

=(150+157+175)/1000

= 482/1000

iii) P(getting no. greater than 4)

= [P(5)+P(6)]/1000= (175+190)/1000

= 365/1000

iv) P(getting multiple of 2)

=[P(2)+P(4)+P(6)]/1000

=(150+149+190)/1000

=489/1000

Hope you have understood

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Answer 2

Answer: 0. 0.179+0.157+0.175

                   =0.511

               ii. 0.15+0.157+0.175

                   =0.482

               iii. 0.175+0.19

                   =0.365

               iv.  0.15 + 0.149 + 0.19

                    = 0.489

Step-by-step explanation:

Prob of outcome 1

179/1000

= 0.179

prob of outcome 2

150/1000

= 0.15

prob of outcome 3

157/1000

= 0.157

prob of outcome 4

149/1000

=0.149

prob of outcome 5

175/1000

=0.175

prob of outcome 190

190/1000

=0.19