A man known to speak the truth 3 out of 4 times. He throws a die and report that it
is six. Find the probability that it is actually six.
Answers
Answered by
8
p(E1)=1/6andP(E)=5/6
Prob.that man speak truth=3/4
not speak truth=1-3/4=1/3
using baye's theorm=
=1/6×3/4/1/6×3/4+5/6×1/4 =3/8
Prob.that man speak truth=3/4
not speak truth=1-3/4=1/3
using baye's theorm=
=1/6×3/4/1/6×3/4+5/6×1/4 =3/8
Answered by
0
Answer:
mark me as brainliest and support me to give more valuable answers.
Step-by-step explanation:
Attachments:
Similar questions