Maths probability chapter extra questions
Answers
Answer:
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Q. 2: A bag contains a red ball, a blue ball and a yellow ball, all the balls being
of the same size. Kritika takes out a ball from the bag without looking into it. What is
the probability that she takes out the
(i) yellow ball?
(ii) red ball?
(iii) blue ball?
Solution:
Kritika takes out a ball from the bag without looking into it. So, it is equally likely that she takes out any one of them from the bag.
Let Y be the event ‘the ball taken out is yellow’, B be the event ‘the ball taken
out is blue’, and R be the event ‘the ball taken out is red’.
The number of possible outcomes = Number of balls in the bag = n(S) = 3.
(i) The number of outcomes favourable to the event Y = n(Y) = 1.
So, P(Y) = n(Y)/n(S) =1/3
Similarly, (ii) P(R) = 1/3
and (iii) P(B) = ⅓answer
Q.3: One card is drawn from a well-shuffled deck of 52 cards. Calculate the probability that the card will
(i) be an ace,
(ii) not be an ace.
Solution:
Well-shuffling ensures equally likely outcomes.
(i) Card drawn is an ace
There are 4 aces in a deck.
Let E be the event ‘the card is an ace’.
The number of outcomes favourable to E = n(E) = 4
The number of possible outcomes = Total number of cards = n(S) = 52 ()
Therefore, P(E) = n(E)/n(S) = 4/52 = 1/13
(ii) Card drawn is not an ace
Let F be the event ‘card drawn is not an ace’.
The number of outcomes favourable to the event F = n(F) = 52 – 4 = 48
Therefore, P(F) = n(F)/n(S) = 48/52 = 12/13 answer
Q.4: Two dice are numbered 1, 2, 3, 4, 5, 6 and 1, 1, 2, 2, 3, 3, respectively. They are thrown and the sum of the numbers on them is noted. Find the probability of getting each sum from 2 to 9 separately.
Solution:
Number of total outcome = n(S) = 36
(i) Let E1 be the event ‘getting sum 2’
Favourable outcomes for the event E1 = {(1,1),(1,1)}
n(E1) = 2
P(E1) = n(E1)/n(S) = 2/36 = 1/18
(ii) Let E2 be the event ‘getting sum 3’
Favourable outcomes for the event E2 = {(1,2),(1,2),(2,1),(2,1)}
n(E2) = 4
P(E2) = n(E2)/n(S) = 4/36 = 1/9
(iii) Let E3 be the event ‘getting sum 4’
Favourable outcomes for the event E3 = {(2,2)(2,2),(3,1),(3,1),(1,3),(1,3)}
n(E3) = 6
P(E3) = n(E3)/n(S) = 6/36 = 1/6
(iv) Let E4 be the event ‘getting sum 5’
Favourable outcomes for the event E4 = {(2,3),(2,3),(4,1),(4,1),(3,2),(3,2)}
n(E4) = 6
P(E4) = n(E4)/n(S) = 6/36 = 1/6
(v) Let E5 be the event ‘getting sum 6’
Favourable outcomes for the event E5 = {(3,3),(3,3),(4,2),(4,2),(5,1),(5,1)}
n(E5) = 6
P(E5) = n(E5)/n(S) = 6/36 = 1/6
(vi) Let E6 be the event ‘getting sum 7’
Favourable outcomes for the event E6 = {(4,3),(4,3),(5,2),(5,2),(6,1),(6,1)}
n(E6) = 6
P(E6) = n(E6)/n(S) = 6/36 = 1/6
(vii) Let E7 be the event ‘getting sum 8’
Favourable outcomes for the event E7 = {(5,3),(5,3),(6,2),(6,2)}
n(E7) = 4
P(E7) = n(E7)/n(S) = 4/36 = 1/9
(viii) Let E8 be the event ‘getting sum 9’
Favourable outcomes for the event E8 = {(6,3),(6,3)}
n(E8) = 2
P(E8) = n(E8)/n(S) = 2/36 = 1/18 answer
Q.5: A coin is tossed two times. Find the probability of getting at most one head.
Solution:
When two coins are tossed, the total no of outcomes = 22 = 4
i.e. (H, H) (H, T), (T, H), (T, T)
Where,
H represents head
T represents the tail
We need at most one head, that means we need one head only otherwise no head.
Possible outcomes = (H, T), (T, H), (T, T)
Number of possible outcomes = 3
Hence, the required probability = ¾ answer
Q.6: An integer is chosen between 0 and 100. What is the probability that it is
(i) divisible by 7?
(ii) not divisible by 7?
Solution:
Number of integers between 0 and 100 = n(S) = 99
(i) Let E be the event ‘integer divisible by 7’
Favourable outcomes to the event E = 7, 14, 21,…., 98
Number of favourable outcomes = n(E) = 14
Probability = P(E) = n(E)/n(S) = 14/99
(ii) Let F be the event ‘integer not divisible by 7’
Number of favourable outcomes to the event F = 99 – Number of integers divisible by 7
= 99-14 = 85 answer
Hence, the required probability = P(F) = n(F)/n(S) = 85/99
Q. 7: If P(E) = 0.05, what is the probability of ‘not E’?
Solution:
We know that,
P(E) + P(not E) = 1
It is given that, P(E) = 0.05
So, P(not E) = 1 – P(E)
P(not E) = 1 – 0.05
∴ P(not E) = 0.95answer
Q. 8: 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just
look at a pen and tell whether or not it is defective. One pen is taken out at random from
this lot. Determine the probability that the pen is taken out is a good one.
Solution:
Numbers of pens = Numbers of defective pens + Numbers of good pens
∴ Total number of pens = 132 + 12 = 144 pens
P(E) = (Number of favourable outcomes) / (Total number of outcomes)
P(picking a good pen) = 132/144 = 11/12 = 0.916answer
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