1
You visited us
1
times! Enjoying our articles?
Unlock Full Access!
Byju's Answer
Standard XII
Mathematics
Odds in Favor and Odds against an Event
Two numbers a...
Question
Two numbers are selected at random from
1
,
2
,
3
,
.
.
.
.
.100
and multiplied. Then find the probability that the product thus obtained is divisible by
3
.
Open in App
Solution
Numbers from
1
,
2
,
,
.
.100
not divisible by
3
are
100
−
33
=
67
Probability that no number is divisible by
3
is
67
C
2
100
C
2
=
0.44666
Then required probability is
1
−
0.44666
=
055
Suggest Corrections
0
Similar questions
Q.
Two numbers are selected at random from 1,2,3,..., 100 and multiplied. Find the probability correct to two places of decimals that the product thus obtained, is divisible by 3.
Q.
Two numbers are selected at random from
1
,
2
,
3
,
.
.
.
.
,
100
and are multiplied, then the probability (correct to two places of decimals) that the product thus obtained is divisible by
3
is
Q.
Two numbers are selected at random from 1, 2, 3,..... , 100 & are multiplied . Find the probability correct to two places of decimals that the product thus obtained, is divisible by 3.
Q.
Two distinct numbes are randomly
chosen from the set
{1,2,3,....,100}
and multiplied together. The probability that this product is divisible by 7 is
Q.
Two numbers
x
and
y
are chosen at random (without replacement) from amongst the numbers
1
,
2
,
3......3
n
. Then find the probability that
x
3
+
y
3
is divisible by
3
.
View More
Join BYJU'S Learning Program
Grade/Exam
1st Grade
2nd Grade
3rd Grade
4th Grade
5th Grade
6th grade
7th grade
8th Grade
9th Grade
10th Grade
11th Grade
12th Grade
Submit
Related Videos
The Immune System
MATHEMATICS
Watch in App
Explore more
Odds in Favor and Odds against an Event
Standard XII Mathematics
Join BYJU'S Learning Program
Grade/Exam
1st Grade
2nd Grade
3rd Grade
4th Grade
5th Grade
6th grade
7th grade
8th Grade
9th Grade
10th Grade
11th Grade
12th Grade
Submit
AI Tutor
Textbooks
Question Papers
Install app