Math, asked by AlishKoiz, 1 year ago

if three lines are drawn on a white board at random with the eyes closed ,show that the probability of forming a triangle is 1/7​


RanaNo: No of zero vertex =1. No of one vertex=4. No of two vertex=3. No of three vertex=1. Reqd prob = 1/9
ushamadhityaual: May i answer the question
ushamadhityaual: Or am i too late for the answer

Answers

Answered by Ankitkumar200314
0

The Required answer is here....

3/21 = 3 ÷ 21

= 1 ÷ 7

3/21 = 1/7

It can be divided further as

3/21 = 1/7

= 1 ÷ 7

3/21 = 0.142......

Hope it helps you.....

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Ankitkumar200314: plzzzz
AlishKoiz: i will mark u as MF
Ankitkumar200314: What's MF???
Randomboy123: I will also mark as MF
Ankitkumar200314: bhai MF kya h???
Randomboy123: Google pe dekh le
Randomboy123: Bhagwan teri ma ko bhi de
ushamadhityaual: All are wrong guys
Answered by ushamadhityaual
3

Answer:

Step-by-step explanation:

Let AB CD and EF be the three lines.Then there are 7 possible cases.

Case1..... The 3 lines coincide each other

Case2...... The 3 lines pass through a same point ie concurrent lines

Case 3......The 3 lines are parallel to each other.

Case4...... AB and CD are parallel and EF is not

parallel to any of them

Case5.....CD and EF are parallel but AB is not parallel to any of them

Case6..... EF and AB are parallel but CD is not parallel to any of them

Case 7......The 3 lines are not mutually parallel to each other ie form a triangle

Here, no. of favourable outcome=1

No. of possible outcomes=7

:. The required probability is 1/7

Hence shown...

Hope you like the answer

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