Science and Technology Workbook for Class X 37
Q.1 In your own words, explain how can you find the conditional probability P(E/ F). (4 marks)
1. Let n be the total number of equally likely outcomes of a random experiment of which m outcomes are
favourable to the event F.
Let the number of outcomes favourable to both E and F be m1. Then m1 outcomes favourable to both E
and F are included in the m outcomes favourable to event F.
P(E ∩ F) = The probability that E and F will happen simultaneously.
P(E ∩ F) =
P(E ∩ F) = ⋅ ...(1)
Now, = Probability of E when the event F has already occurred. = P(E | F)
P(E ∩ F)
Putting these values in (1), we get P(E ∩ F) = P(E|F).P(F) ⇒ P(E | F) =
Q.2 Your friend asks you to explain how the product rule for independent events differs from the product rule
for dependent events. How would you respond? (4 marks)
2. If two events E and F are such that, occurrence or non-occurrence of any one of the events does not affect
the Probability of occurrence or non-occurrence of the other then the two events are called independent
events otherwise dependent.
In case of independent events the product rule is P(E ∩ F) = P(E) × P(F)
In case of dependent event the product rule is P(E ∩ F) = P (E | F) ⋅ P(F) , provided P ( F ) ≠ 0
Or P(E ∩ F) = P (F | E) ⋅ P(E) provided P ( E ) ≠ 0
Q.3 A scientist wishes to determine whether there is any dependence between color blindness (C) and
deafness (D). Given the probabilities listed in the table, what should his findings be ? (4 marks)
D D' Totals
C .0004 .0796 .0800
C' .0046 .9154 .9200
Totals .0050 .9950 1.0000
3. Events colour blindness (C) and deafness (D) will be dependent if P(C|D) ≠ P(C)
n(C ∩ D)
P(C ∩ D) n(S) n(C ∩ D) 0.0004 2
Now, P(C|D) = = n(D) = = = = 0.08
P(D) n(D) 0.0050 25
P(C) = .0800 = .08
Hence, P(C|D) = P(C) Hence, events C and D are independent events.
38 Science and Technology Workbook for Class X
Q.4 An urn contains 5 white and 8 black balls. Two successive drawings of three balls at a time are made such
that the balls are not replaced before the second draw. Find the probability that the first draw gives 3 white
balls and second draw gives 3 black balls. (4 marks)
4. Consider the following events:
A = drawing 3 white balls in first draw, B = drawing 3 black balls in the second draw.
Required probability = P(A ∩ B) = P(A)P(B|A) ...(i)
C3 5! 3!10! 5
Now, P(A) = = × =
13 3!2! 13! 143
After drawing 3 white balls in first draw 10 balls are left in the bag, out of which 8 are black balls.
C3 8! 3!7! 7
∴ P(B|A) = = × =
10 3!5! 10! 15
Hence, Required probability = P(A ∩ B) = P(A). P (B | A) [From (i)]
5 7 7
= × =
143 15 429
Q.5 The Motor Vehicle Department has found that the probability of a person passing the test for a driver's
license on the first try is .75. The probability that an individual who fails on the first test will pass on the
second try is .80, and the probability that an individual who fails the first and second tests will pass the
third time is .70. Find the probability that an individual. (1 + 1 + 2) marks
(i) fails both the first and second tests. (ii) will fail first three tests
(iii) will require at least two tries to pass the test.
5. (i) P(fails both the first test and second test ) = P(fails 1st test) × P(fails 2nd test)
= (1 – 0.75)(1 – 0.80) = 0.25 × 0.20 = 0.0500 = 0.05
(ii) P(will fail three times in a row) = P(fails 1st test) × P(fails 2nd test) × P(fails 3rd test)
= (1 – 0.75) (1 – 0.80) (1 – 0.70) = 0.25 × 0.20 × 0.30 = 0.015
(iii) P(will require at least two tries to pass the test)
= P(Pass in two tries) + P(Pass in three tries)
= P(fails in the first try and passes in the 2nd try) + P(fails in the first try, fails in 2nd try and passes
in 3rd try)
= (1 – 0.75) (0.80) + (1 – 0.75) (1 – 0.80) (0.70) = 0.25 × 0.80 + 0.25 × 0.20 × 0.70 = 0.2 + 0.035 = 0.235.