Question

A die is loaded such that 1 turning upwards is 2 times as often as 6 and 3 times as any other face (2 or 3 or 4 or 5). The probability that we get a face with 6 when we throw such a die is​

Answer 1

Hey Buddy..

Here's the answer...

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# For denotation...P(x) = probability of x

=> A.T.Q.

P(1) = 2 P(6)

AND

P(1) = 3 P(2)

P(1) = 3 P(3)

P(1) = 3 P(4)

P(1) = 3 P(5)

# Suppose probability of 1 is K , i.e P(1) = K

=> P(1) = K

=> P(2) = K / 3

=> P(3) => K / 3

=> P(4) = K / 3

=> P(5) = K / 3

=> P(6) = K / 2

# We know some of all probability is one

=> P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1

=> K + ( K/3 ) + ( K/3 ) + ( K/3 ) + ( K/3 ) + K / 2 = 1

=> K + ( 4 K/3 ) + K/2 = 1

=> ( 6 K + 8 K + 3 K )/6 = 1

=> 17 K = 6

=> K = 6 / 17

# We have to find probability of 6 , i.e P( 6 )

=> P(6) = K/2

=> P(6) = ( 6/17 ) × ( 1/ 2 )

=> P(6 ) = 3 / 17

HOPE HELPED..

JAI HIND

:)