Probability
Conditional probability, independent events and binomial chances.
Learn with Kutty
Go one screen at a time. Press Next when you are ready. Tap 🔊 Read to me to listen.
If you already know something happened, only those cases count.
Look: Machine A makes 60% of bolts (2% faulty). B makes 40% (5% faulty).
✏️ Your turn!
Type your answer and press Check. Kutty will tell you if it is right.
1A dice shows an even number. Chance it is 6?
2P(A) = 0.5, P(B) = 0.4, independent. Both?
3P(both) = 0.12, P(B) = 0.3. P(A given B)?
! Careful!
✗Answering 0.05.
✓That is "faulty given B". We want "B given faulty".
✗Treating cards drawn without putting back as independent.
✓After one is drawn, fewer cards remain.
See the whole lesson on one page
If you already know something happened, only those cases count.
Chance of A given B = (both happen) ÷ (B happens). A tree diagram helps: multiply along branches, add across them.
Think of it like thisIf you know it is cloudy, the chance of rain is higher than on a random day. Knowing something changes the odds.
Solved together
A bolt is faulty. What is the chance it came from machine B?
Multiply along each faulty branch.
A and faulty: 0.6 × 0.02 = 0.012B and faulty: 0.4 × 0.05 = 0.020All faulty bolts together.
0.012 + 0.020 = 0.032B’s share of the faulty ones.
0.020 ÷ 0.032 = 0.625
Answer0.625 (5 out of 8).
For older learners and grown-ups: a little more
P(A|B) = P(A ∩ B)/P(B). Bayes’ theorem: P(B|F) = P(B)P(F|B)/P(F). Independent events: P(A ∩ B) = P(A)P(B).
Now practise it
Kanakku makes new questions on probability every time. It shows the working and gives a hint when you are stuck.