Lessons 34–35 · 2 lessons · M. A. Mirzaahmedov, Sh. N. Ismailov, A. Q. Amanov (algebra and analysis), B. Q. Haydarov (geometry). Mathematics Grade 11, Parts 1 and 2, 1st edition. ZAMIN NASHR, Tashkent, 2018
35
Adding and multiplying probabilities; methods of computing probabilities
Textbook, Part 2: pp. 83–90
GoalLearn to apply the addition rule, conditional probability and the multiplication rule in probability problems.
New words
conditional probability · shartli ehtimollikindependent events · bog‘liq bo‘lmagan hodisalaraddition rule · qo‘shish qoidasimultiplication rule · ko‘paytirish qoidasi
Explanation
Addition rule: P(A ∪ B) = P(A) + P(B) − P(A ∩ B); for mutually exclusive events the last term is zero. The probability of A given that B has occurred is the conditional probability P(A|B) = P(A ∩ B)/P(B). Hence the multiplication formula P(A ∩ B) = P(B) · P(A|B). If P(A|B) = P(A), then A and B are independent and P(AB) = P(A) · P(B); then A and B̄, and Ā and B̄, are also independent. When drawing from a bag without replacement each later probability depends on the earlier result: for example P(both white) = (5/8) · (4/7). The probability that at least one of independent events occurs is P = 1 − (1 − p₁)(1 − p₂)…(1 − pₙ). Do not confuse being mutually exclusive with being independent: two independent events with positive probabilities can occur together.
Worked examples
In a class 40 % of the students go swimming, 30 % play chess and 12 % do both. At least one of the two: P = 0.4 + 0.3 − 0.12 = 0.58. P(swimming | chess) = 0.12/0.3 = 0.4 = P(swimming), so these events are independent. A bag has 5 white and 3 black balls; two are drawn without replacement: P(both white) = 5/8 · 4/7 = 20/56 = 5/14.
Two students solve a problem independently, with probabilities 0.7 and 0.6. Both solve it: 0.7 · 0.6 = 0.42. Exactly one solves it: 0.7 · 0.4 + 0.3 · 0.6 = 0.28 + 0.18 = 0.46. At least one: 1 − 0.3 · 0.4 = 0.88.
Class activity
“With and without replacement”: put 3 red and 2 blue pencils in a bag. Draw two pencils (a) with replacement, (b) without replacement, 30 times each, and record the frequency of both red. Compare with the theoretical values (3/5)² = 9/25 and 3/5 · 2/4 = 3/10.
Practice
1
A and B are independent, P(A) = 0.3, P(B) = 0.5. Find P(AB).
0.15
2
A and B are mutually exclusive, P(A) = 0.4, P(B) = 0.3. Find P(A ∪ B).
0.7
3
A bag holds 6 red and 4 green balls. Two are drawn without replacement. What is the probability that both are green?
2/15
4
Why can independent events with positive probabilities not be mutually exclusive?
For independent events P(AB) = P(A)P(B) > 0, so they can occur together. For mutually exclusive events P(AB) = 0.