# .    The following table gives distance (in km) that 40 engineers have to travel from their residences to their work places:-   Distance (in km) 0 - 5 5 - 10 10 - 15 15 - 20 20 - 25 25 - 30 30 - 35 No of engineers 5 11 11 9 1 1 2  Find the probability that an engineer selected at random lives at a distance of:-(i) 10 – 15 km (event E1)  (ii) more than 35 km (event E2)       (iii) less than 10 km (event E3)        (iv) upto 35 km answers in step please

Dear Student,

We have the following table gives distance (in km) that 40 engineers have to travel from their residences to their work places:

 Distance (in km) 0 - 5 5 - 10 10 - 15 15 - 20 20 - 25 25 - 30 30 - 35 No. of engineers 5 11 11 9 1 1 2

We know : Probability P ( E )  =

Here ,

n ( S ) = 40

i ) 10 - 15 km ( event E1 )  ,  From given table we get

n ( E1 ) = 11

Therefore,

Probability that an engineer selected at random lives at a distance of ( 10 - 15 km ) = $\frac{\mathbf{11}}{\mathbf{40}}$                             ( Ans )

ii ) more than 35 km ( event E2 )  ,  From given table we get

n ( E2 ) = 0

Therefore,

Probability that an engineer selected at random lives at a distance of ( more than 35 km ) = $\frac{0}{40}$ = 0                             ( Ans )     iii ) less than 10 km (event E3)  ,  From given table we get

n ( E3 ) = 5 +  11 =  16

Therefore,

Probability that an engineer selected at random lives at a distance of ( less than 10 km ) =                             ( Ans )  iv ) Upto 35 km ,  From given table we get

n ( E4 ) = 5 + 11 + 11 + 9 + 1 + 1 + 2 =  40

Therefore,

Probability that an engineer selected at random lives at a distance of ( upto 35 km ) = $\frac{40}{40}$ = 1                             ( Ans )

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