Probability
Basic Knowledge
Probability(m¤¢ve¨Zv): †Kvb GKwU NUbv NUvi m¤¢vebv‡K Probability(m¤¢ve¨Zv) e‡j|
m~Ît m¤¢ve¨Zv = D³ NUbvi AbyK‚j msL¨v / ‡gvU bgybv msL¨v
†R‡b ivL‡Z n‡et
1| †h NUbvwU Aek¨B NU‡e Zvi m¤¢ve¨Zv gvb 1 Ges †h NUbvwU NUvi †Kvb m¤¢vebv †bB, Zvi m¤¢ve¨Zv
gvb 0(k~b¨)| A_©vr m¤¢ve¨Zv mxgv 0 †_‡K 1|
2| K‡q‡bi †¶‡Ît GKwU K‡q‡bi `ywU wcV Av‡Q(Head and Tail)
K‡qbwU 1 evi wb‡¶c Ki‡j Head A_ev Tail co‡Z cv‡i| A_©vr †gvU bgybv msL¨v = 2 wU
GKB K‡qb evi evi wb‡¶‡ci †¶‡Ît K‡qb msL¨v n wU n‡j †gvU bgybv msL¨v = 2n n‡e|
3| GKwU Q°vi 6wU wcV _v‡K| hv‡Z 1,2,3,4,5,6 msL¨v wPwýZ _v‡K|
GKwU Q°v GK evi wb‡¶c Ki‡j, †gvU bgybv †¶Î = 6 wU
GKB Q°v evi evi (n evi) wb‡¶c Ki‡j, †gvU bgybv msL¨v = 6n wU
4| GKwU Zv‡mi c¨v‡K‡U †gvU 52 wU KvW© _v‡K| GB 52 wU KvW© †gŠwjK 4wU fv‡M wef³|
h_vt niZb, iæBZb, B¯‹vcb I wPov| Giv cÖ‡Z¨‡K Avevi 13 wU fv‡M wef³|
h_vt ivRv, ivbx, †U°v, †Kv‡Zvqvj, 2,3,4,5,6,7,8,9,10|
A_©vr 52wU Kv‡W©i g‡a¨ 4wU ivRv, 4 wU ivbx, 4 wU †U°v BZ¨vw` Av‡Q|
ZvQvov 52wU KvW© Avevi 2 fv‡M wef³| h_vt jvj I Kv‡jv|
A_©vr 26 wU jvj I 26 wU Kv‡jv KvW© Av‡Q|
5| GKwU wbw`©ó NUbv bv NUvi m¤¢ve¨Zv = 1 - NUbv NUvi m¤¢ve¨Zv
Probability
Learning Section
1. 3 coins are tossed at random. Show the sample space and find the probability of
getting:
(i) One head and two tails
(ii) One tail
(iii) One tail and two heads [BB AD 18, HBFC SO 17]
঄নুবাদঃ
তিনতি মুদ্রা দদবভাবব তনবেপ করা হ঱। নমুনা ক্ষেত্রতি ত঱খুন এবং তনম্ন ত঱তখি ঘিনা সমূবহর সম্ভাবিা তনর্ণয় করুনঃ (i) একতি
ক্ষহড এবং দুতি ক্ষিআ঱ (ii) একতি ক্ষিআ঱ (iii) একতি ক্ষিআ঱ এবং দুতি ক্ষহড
Solution:
If 3 coins are tossed, Probable sample space
= {HHH, HHT,HTH,THH, TTT, TTH, THT, HTT}. So, the number of total outcomes = 8
(i) Probability of getting one head and two tails:
From sample space we get TTH,THT and HTT.
So, number of favorable outcomes = 3
Required Probability =
=
(ii) Probability of getting one tail:
From sample space we get HHT, HTH and THH.
So, number of favorable outcomes = 3
Required Probability =
=
(iii) Probability of getting one tail and two heads:
From sample space we get HHT, HTH and THH.
So, number of favorable outcomes = 3
Required Probability =
=
2. A bag contains 75 rods. 35 are blue and 25 of these blue rods are twisted at the
bottom. The rest of them are red and 30 of the red ones are twisted. The rods that are
not twisted are clear. What is the probability of drawing? [Combined 2 Banks SO(IT) 18]
a. A blue rod from the box
b. A clear rod from the box
c. A blue, twisted rod
d. A red, clear rod
e. A twisted rod
঄নুবাদঃ
একতি বযাবগ ৭৫ তি ঱াতি রবয়বে। এবদর মবযয ৩৫ তি নী঱ এবং নী঱ ঱াতিগুব঱ার মবযয ২৫তির অগায় বাাঁকাবনা। বাতক ঱াতি গুব঱া
঱া঱ এবং এবদর মবযয ৩০তি ঱াতি বাাঁকাবনা। ক্ষয ঱াতিগুব঱া বাাঁকাবনা নয়, ক্ষসগুব঱া ক্ষসাজা। ঱াতি ক্ষববে ক্ষনওয়ার সম্ভাবযিা কি?
ক) বক্স ক্ষেবক একতি নী঱ ঱াতি খ) বক্স ক্ষেবক একতি ক্ষসাজা ঱াতি
গ) একতি বাাঁকাবনা নী঱ ঱াতি
ঘ) একতি ক্ষসাজা ঱া঱ ঱াতি
ঙ) একতি বাাঁকাবনা ঱াতি
Solution:
The number of rods = 75, Blue rods = 35 and Red rods = 75-35 = 40
Twisted Blue rods = 25 and Clear Blue rods = 35-25 = 10
Exam Aid Bank Written Math (4th Edition)
Page 96
Probability
Twisted Red rods = 30 and Clear Red rods = 40-30 = 10
Total clear rods = 10+ 10 = 20 and total twisted rods = 25 + 30 = 55
a) The probability of drawing a Blue rod from the box = = (Ans.)
b) The probability of drawing a Clear rod from the box =
c) The probability of drawing a Blue, twisted rod =
=
(Ans.)
= (Ans.)
d) The probability of drawing a Red, clear rod from the box =
e) The probability of drawing a Twisted rod from the box =
=
=
(Ans.)
(Ans.)
3. A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random.
What is the probability that none of the balls drawn is blue?
Solution:
Total balls = 2+3+2 = 7
Number of ways drawing 2 balls from 7 balls = 7C2 = 21
Number of ways drawing 2 not blue balls = 5C2 = 10
The probability that none of the balls drawn is blue = (Ans.)
Alternative Method:
Total balls = 2+3+2 = 7
Probability of drawing not blue ball =
Remaining total balls = 7-1 = 6 and not blue balls = 5-1 = 4
Probability of drawing not blue ball = =
Required probability = × =
(Ans.)
4. A box has 6 black, 4 red, 2 white and 3 blue shirts. What is the probability that 2
red shirts and 1 blue shirt get chosen during a random selection of 3 shirts from the
box?
Solution:
Total = 6 + 4 + 2 + 3 = 15 shirts
Probability of choosing,
(i) blue shirt 1st, then red shirt and then red shirt = × × =
(ii) red shirt 1st, then red shirt and then blue shirt =
× ×
=
(iii) red shirt 1st, then blue shirt and then red shirt =
× ×
=
∴ Total probability =
+
+
=
(Ans.)
Exam Aid Bank Written Math (4th Edition)
Page 97
Probability
5. A bag contains 6 red balls, 11 yellow balls and 5 pink balls. If two balls are drawn
at random from the bag. One after another what is the probability that the first ball
is red and second ball is yellow?
Solution:
Number of red balls = 6, yellow balls = 11 and pink balls = 5
Total number of balls = 6 + 11 + 5 = 22
Probability of the first ball red = =
Probability of the second ball yellow =
∴ Required Probability =
×
= (Ans.)
6. A box contains 3 blue marbles, 4 red, 6 green marbles and 2 yellow marbles. If two
marbles are drawn at random, what is the probability that at least one is green?
[Rupali SO 19]
Solution:
Given that,
3 blue marbles, 4 red, 6 green marbles and 2 yellow.
Total = 3+4+6+ 2 = 15
[Green = 6 and non-green = 3+4+2 = 9]
Probability that at least one green marble drawing of two marbles
= Probability that one is green + Probability that both are green
=
+
=
+
=
=
= (Ans.)
7. An article manufactured by a company consists of two parts A and B. In the
process of manufacture of part A, 9 out of 100 are likely to be defective. Similarly, 5
out of 100 are likely to be defective in the process of manufacture of part B. The
probability that the assembled part will not be defective is- [Sonali FF SO 19, Combined 5
Banks Cash 19]
Abyev`t
GKwU †Kv¤cvbx GKwU `ª‡e¨i A I B Ask cÖ¯Z
‘ K‡i| A Ask cÖ¯Z
‘ Ki‡Z wM‡q 100wUi g‡a¨ 9 wU ÎæwUc~Y©
†_‡K hvq Ges B As‡ki 100 wUi g‡a¨ 5 wU ÎæwUc~Y© †_‡K hvq| A I B Ask GK‡Î ÎæwUc~Y© bv nIqvi m¤¢ve¨Zv
KZ?
Solution:
†bvUt m¤¢ve¨Zvi †¶‡ÎGiven that, Part A has 9 defective articles out of 100.
 A I B Gi m¤¢ve¨Zv †ei
So, non-defective articles in part A = 100-9 = 91
Ki‡Z ej‡j Df‡qi cÖvß
Probability of non-defective articles in part A =
m¤¢ve¨Zv ¸Y Ki‡Z n‡e|
 A A_ev B Gi m¤¢ve¨Zv †ei
Part B has 5 defective articles out of 100.
Ki‡Z ej‡j Df‡qi cÖvß
So, non-defective articles in part B = 100-5 = 95
m¤¢ve¨Zv †hvM Ki‡Z n‡e|
Probability of non-defective articles in part B =
Probability of non-defective articles in assembles part
= × = × =
= 0.8645 (Ans.)
Exam Aid Bank Written Math (4th Edition)
Page 98
Probability
8. A man and his wife appear in an interview for two vacancies in the same post. The
probability of husband's selection is and the probability of wife's selection is .
What is the probability that only one of them is selected?
Solution:
Given that, the probability of husband's selection = and wife's selection =
The probability of husband's not selection = 1- =
The probability of wife's not selection = 1- =
The probability of selecting husband and not selecting wife = × =
The probability of selecting wife and not selecting husband = × =
The probability that only one of them is selected =
+
=
= (Ans.)
9. Two cards are drawn at random from a pack of 52 cards. What is the probability
that either both are black or both are queens?
Solution:
Total cards = 52, black cards = 26 and queen cards = 4
Probability that both cards are black =
=
Probability that both cards are queen =
=
Probability that counted both cases =
Required probability =
+
=
-
=
=
=
(Ans.)
10. A box contains 3 blue marbles, 4 red, 6 green marbles and 2 yellow marbles. If
two marbles are picked at random, what is the probability that they are either blue
or yellow?
Solution:
Given that, 3 blue marbles, 4 red, 6 green marbles and 2 yellow marbles.
Total = 3+4+6+2 = 15 marbles
Probability that both marbles are blue =
=
=
Probability that both are yellow =
=
Probability that one blue and other is yellow =
Required probability =
+
+
=
=
=
=
=
(Ans.)
11. A box contains 6 bottles of variety 1 drink, 3 bottles of variety 2 drink and 4
bottles of variety 3 drinks. Three bottles of them are drawn at random, what is the
probability that the three are not of the same variety. [Rupali & Janata Officer 20]
Exam Aid Bank Written Math (4th Edition)
Page 99
Probability
Solution:
Total number of drink bottles = 6 + 3 + 4 = 13
Probability of getting variety 1 drink =
=
Probability of getting variety 2 drink =
=
Probability of getting variety 3 drink =
=
Probability of getting same color balls =
+
+
Probability of getting not of the same color = 1 -
=
=
(Ans.)
Explanation: প্রেবম একআ যরবর্র তরংক গুব঱ার সম্ভাবিা অ঱াদা অ঱াদাভাবব ক্ষবর করবি হবব িারপর
ক্ষমাি(১)ক্ষেবক সব গুব঱ার ক্ষযাগফ঱ তববয়াগ করব঱,‘একআ যরবর্র নয়’ িার সম্ভাবযিা পাওয়া যাবব।
12. A bag contains some white and black balls. The probability of picking two white
balls one after other without replacement from that bag is 14/33. Then what will be
the probability of picking two black balls from that bag if bag can hold maximum 15
balls only? [Combined 7 Banks SO 18, Combined 5 Banks AME/HE/AE(IT) 18]
Abyev`t
GKwU e¨v‡M wKQz mv`v I Kv‡jv ej i‡q‡Q| e`jv‡bv Qvov cici `yBwU ej Zzj‡j `yBwU ejB mv`v nevi m¤¢ve¨Zv
14/33| Zvn‡j cici `yBwU ej Zzj‡j `yBwU ejB Kv‡jv nevi m¤¢ve¨Zv KZ hw` e¨v‡M m‡e©v”P 15wU ej _v‡K?
Solution:
Let, the number of white balls be x and black balls be y.
According to the question,
(
=
)
(
Or,
[Probability =
)
(
(
Or,
Or,
Or, (
) (
(
)
(
(
Formula nCr = (𝑛
=
)
)
)
)
×(
(
)(
]
×(
)(
)
)
=
(
)(
)
)
)(
=
)
𝑛
𝑟) 𝑟
=
[Multiplying the denominator and numerator by 4]
`ywU µwgK msL¨vi ¸Ydj evbv‡bvi Rb¨ je I ni‡K 4 w`‡q ¸Y Kiv n‡q‡Q
Here, clearly x, (x-1) and (x+y), (x+y-1) both are consecutive numbers.
So, we can say the possible value of x = 8 and (x+y) = 12
y = 12-8 = 4
The number of white balls = 8, black balls = 4 and total = 12 balls. (not more than 15)
Required probability =
= = (Ans.)
Exam Aid Bank Written Math (4th Edition)
Page 100
Probability
13. A bag contains 10 white and 15 black balls. Two balls are drawn in succession.
What is the probability that first is white and second is black? [Combined 7 Banks SO 21]
Solution:
Number of white balls = 10 and black balls = 15
Total number of balls = 10 + 15 = 25
Probability of the first ball white = =
Probability of the second ball black =
= [একতি ব঱ িুব঱ তনব঱ ২৫-১ = ২৪তি ব঱ োকবব]
∴ Required Probability = × = (Ans.)
Important questions for practice
1. A box contains 5 green, 4 yellow and 3 white balls. Three balls are drawn at
random. What is the probability that they are not of the same colour? [BASIC Bank AM 18]
2. A bag contains 5 red and 3 green balls. Another bag contains 4 red and 6 green
balls. If one ball is drawn from each bag. Find the probability that one ball is red and
one is green. [PKB EO 19]
3. First bag contains 4 red and 3 black balls. Second bag contains 2 red and 4 black
balls. One bag is selected at random. From the selected bag, one ball is drawn. Find
the probability that the ball drawn is red. [Combined 4 Banks Officer 19]
4. A brother and sister appear for an interview against two vacant posts in an office.
The probability of the brother's selection is and that of the sister's selection is .
What is the probability that one of them is selected?
5. There are two bags A and B. A contains 6 red flowers and 3 pink flowers, whereas
bag B contains 2 red flowers and 7 pink flowers. One flower is chosen from a bag
randomly. What is the probability that the flower chosen is pink?
6. Bag x contains 3 red and 5 black balls and bag y contains 4 red and 4 black balls.
One bag is selected at random and from the selected bag one ball is drawn. What is
the probability that the ball drawn is red?
7. Tinu and Pintu go for an interview for two vacancies. The probability for the
selection of Tinu is and whereas the probability for the selection of Pintu is . What
is the probability that none of them are selected?
Exam Aid Bank Written Math (4th Edition)
Page 101
Probability
8. In a box, there are 10 apples and th of the apples are rotten. If three apples are
taken out from the box, what will be the probability that at least one apple is rotten.
[BB AD (Research) 19]
9. A bag contains 6 white and 4 black balls. 2 balls are drawn at random. Find the
probability that they are of same colour.
10. A bag contains 8 white balls, and 3 blue balls. Another bag contains 7 white, and
4 blue balls. What is the probability of getting blue ball?
11. A box contains 3 blue marbles, 4 red, 6 green marbles and 2 yellow marbles. If
three marbles are drawn what is the probability that one is yellow and two are red?
12. A man can hit a target once in 3 shots. If he fires 3 shots in succession, find the
probability that he will hit his target.
13. A class consists of 100 students, 25 of them are girls and 75 boys; 20 of them are
rich and remaining poor; 40 of them are fair complexioned. The probability of
selecting a fair complexioned rich girl is14. In a class, there are 15 boys and 10 girls. Three students are selected at random.
The probability that 1 girl and 2 boys are selected, is:
15. A problem is given to three students whose chances of solving it are ,
respectively. What is the probability that the problem will be solved?
and
16. There are 12 boys and 8 girls in a tuition centre. If three of them scored first
mark, then what is the probability that one of the three is a girl and the other two are
boys?
17. A box contains 3 white, 4 red and 7 blue erasers. If five erasers are taken at
random then the probability that all the five are blue color is:
18. A box has 6 black, 4 red, 2 white and 3 blue shirts. When 2 shirts are picked
randomly, what is the probability that either both are white or both are blue?
19. In a drawer there are 4 white socks, 3 blue socks and 5 grey socks. Two socks are
picked randomly. What is the possibility that both the socks are of same color?
20. A bag contains 4 red, 5 yellow and 6 pink balls. Two balls are drawn at random.
What is the probability that none of the balls drawn are yellow?
Exam Aid Bank Written Math (4th Edition)
Page 102
Probability
Answer:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
Solution of practice session
1. A box contains 5 green, 4 yellow and 3 white balls. Three balls are drawn at
random. What is the probability that they are not of the same colour? [BASIC Bank AM 18]
Abyev`t
GKwU ev‡· 5 wU meyR, 4 wU njy` I 2 wU mv`v ej Av‡Q| 3 wU ej •`e¨fv‡e DËjb Kiv nj| 3wU ej GKB iO bv
nIqvi m¤¢veZv KZ?
Solution:
Probability of getting green balls =
=
Probability of getting yellow balls =
Probability of getting white balls =
Probability of getting same color balls =
=
=
+
+
Probability of getting not of the same color = 1 -
=
=
=
(Ans.)
2. A bag contains 5 red and 3 green balls. Another bag contains 4 red and 6 green
balls. If one ball is drawn from each bag. Find the probability that one ball is red and
one is green. [PKB EO 19]
Solution:
Given that,
1st Bag contains 5 red and 3 green ball. Total = 5+3 = 8 balls
2nd Bag contains 4 red and 6 green balls. Total = 4+6 = 10 balls
Probability of getting one Red ball from 1st bag and one Green from 2nd Bag = × =
Probability of getting one Green ball from 1st bag and one Red from 2nd Bag = ×
Exam Aid Bank Written Math (4th Edition)
=
Page 103
Probability
The probability that one ball is red and one is green = +
=
=
(Ans.)
3. First bag contains 4 red and 3 black balls. Second bag contains 2 red and 4 black
balls. One bag is selected at random. From the selected bag, one ball is drawn. Find
the probability that the ball drawn is red. [Combined 4 Banks Officer 19]
Solution:
Given that,
First bag contain 4 red and 3 black balls; total balls = 7
Second bag contain 2 red and 4 black balls; total balls = 6
Probability of picking 1 red ball from first bag =
Probability of picking 1 red ball from second bag = =
Total = + =
=
and probability of selecting a bag is
Probability of choosing a red ball = ×
=
(Ans.)
Alternative Method:
Probability of getting a red ball taken from first bag = × =
Probability of getting a red ball taken from second bag = × =
Probability of getting a red ball = + =
=
(Ans.)
4. A brother and sister appear for an interview against two vacant posts in an office.
The probability of the brother's selection is and that of the sister's selection is .
What is the probability that one of them is selected?
Solution:
The way brother is selected and sister is not selected = ×(1- ) = × =
The way sister is selected and brother is not selected = ×(1- ) = × =
The probability that one of them is selected =
+
=
=
= (Ans.)
5. There are two bags A and B. A contains 6 red flowers and 3 pink flowers, whereas
bag B contains 2 red flowers and 7 pink flowers. One flower is chosen from a bag
randomly. What is the probability that the flower chosen is pink?
Solution:
Given that,
Bag A contains 6 red and 3 pink flower; total flowers = 9
Bag B contains 2 red and 7 pink flower; total flowers = 9
Probability of picking 1 pink flower from bag A =
Exam Aid Bank Written Math (4th Edition)
Page 104
Probability
Probability of picking 1 pink flower from bag B =
Total = + =
=
And probability of selecting a bag is
Probability of choosing a pink flower = ×
= (Ans.)
6. Bag x contains 3 red and 5 black balls and bag y contains 4 red and 4 black balls.
One bag is selected at random and from the selected bag one ball is drawn. What is
the probability that the ball drawn is red?
Solution:
Probability of selecting a red ball from bag x = probability of selecting bag x ×probability
of selecting red ball from it = × =
Probability of selecting a red ball from bag y = probability of selecting bag y× probability
of selecting red ball from it = × =
The probability of selecting a red ball = probability of selecting a red ball from bag x +
probability of selecting a red ball from bag y.
Probability that the ball drawn is red = + = (Ans.)
7. Tinu and Pintu go for an interview for two vacancies. The probability for the
selection of Tinu is and whereas the probability for the selection of Pintu is . What
is the probability that none of them are selected?
Solution:
Probability of Tinu being selected =
Probability of Tinu not being selected = (1- ) =
Probability of Pintu being selected =
Probability of Pintu not being selected = (1- ) =
Probability that both are not selected = × =
(Ans.)
8. In a box, there are 10 apples and th of the apples are rotten. If three apples are
taken out from the box, what will be the probability that at least one apple is rotten.
[BB AD (Research) 19]
Solution:
The number of rotten apple = ×10 = 4. So, good apple = 10-4 = 6
Probability that at least one of the apples is rotten
= 1 - Probability that none of the bulb is defective = 1 = 1Exam Aid Bank Written Math (4th Edition)
= 1- = (Ans.)
Page 105
Probability
9. A bag contains 6 white and 4 black balls. 2 balls are drawn at random. Find the
probability that they are of same colour.
Solution:
Total balls = 6+4 = 10
Number of ways selecting 2 balls from 10 = 10C2 = 45
Number of ways selecting 2 balls same colour = 6C2+ 4C2 = 15+6 = 21
The probability that they are of same colour = = (Ans.)
10. A bag contains 8 white balls, and 3 blue balls. Another bag contains 7 white, and
4 blue balls. What is the probability of getting blue ball?
Solution:
Probability of getting blue ball taken from first bag = × =
Probability of getting blue ball taken from second bag = ×
Probability of getting blue ball =
+
=
=
=
(Ans.)
11. A box contains 3 blue marbles, 4 red, 6 green marbles and 2 yellow marbles. If
three marbles are drawn what is the probability that one is yellow and two are red?
Solution:
Given that, 3 blue marbles, 4 red, 6 green marbles and 2 yellow.
Total = 3+4+6+ 2 = 15
The probability that one is yellow and two are red =
=
=
(Ans.)
12. A man can hit a target once in 3 shots. If he fires 3 shots in succession, find the
probability that he will hit his target.
Solution:
Probability of hitting the target = (1- probability of not hitting the target)
If he hits once every 3 balls, then probability of hitting =
So, probability of not hitting = 1- =
Probability of not hitting the target all three times = × × =
Probability of hitting target = 1 -
=
(Ans.)
13. A class consists of 100 students, 25 of them are girls and 75 boys; 20 of them are
rich and remaining poor; 40 of them are fair complexioned. The probability of
selecting a fair complexioned rich girl isSolution:
The probability of selecting girl =
=
The probability of selecting rich =
=
Exam Aid Bank Written Math (4th Edition)
Page 106
Probability
The probability of selecting fair complexioned =
=
Here, three are independent events.
Probability of rich and fair complexioned girl = × × =
(Ans.)
14. In a class, there are 15 boys and 10 girls. Three students are selected at random.
The probability that 1 girl and 2 boys are selected, is:
Solution:
Number of ways of selecting 1 girl and 2 boys = 10C1×15C2 = 10×105 = 1050
Number ways of selecting 3 students out of 25 = 25C3 = 2300.
Required probability =
= (Ans.)
15. A problem is given to three students whose chances of solving it are ,
respectively. What is the probability that the problem will be solved?
Solution:
Probability that the problem will be solved = 1 – None solve the problem
Probability that 1st student not solve the problem = 1- =
and
Probability that 2nd student not solve the problem = 1- =
Probability that 3rd student not solve the problem = 1- =
Probability that none solve the problem = × × =
Probability that the problem will be solved = 1 – = (Ans.)
16. There are 12 boys and 8 girls in a tuition centre. If three of them scored first
mark, then what is the probability that one of the three is a girl and the other two are
boys?
Solution:
Total number of students = 20
The number of ways of three scored first mark = 20C3 =
= 1140
The number of possible outcomes of 1 girl out of 8 and 2 boys out of 12 = 8C1×12C2
=
= 528
Required probability =
=
(Ans.)
17. A box contains 3 white, 4 red and 7 blue erasers. If five erasers are taken at
random then the probability that all the five are blue color is:
Solution:
Total number of erasers in the box = 3+ 4+ 7 = 14
Exam Aid Bank Written Math (4th Edition)
Page 107
Probability
The number of ways of taking 5 out of 14 = 14C5 =
The number ways of getting all the 5 blue erasers = 7C5 =
Required probability =
=
= 2002
= 21
(Ans.)
18. A box has 6 black, 4 red, 2 white and 3 blue shirts. When 2 shirts are picked
randomly, what is the probability that either both are white or both are blue?
Solution:
Total shirts = 6 + 4+ 2 + 3 = 15
Probability for getting 2 white shirts = × =
Probability for getting 2 blue shirts =
Total Probability =
+
=
×
=
(Ans.)
19. In a drawer there are 4 white socks, 3 blue socks and 5 grey socks. Two socks are
picked randomly. What is the possibility that both the socks are of same color?
Solution:
Total socks = 4 + 3 + 5 = 12
Probability of getting 1 White sock =
Probability of getting 2 White socks =
Probability of getting White socks = × =
Similarly,
Probability of getting Blue socks = × =
Probability of getting Grey socks =
×
∴ Total Probability =
(Ans.)
+
+
=
=
20. A bag contains 4 red, 5 yellow and 6 pink balls. Two balls are drawn at random.
What is the probability that none of the balls drawn are yellow?
Solution:
Number of red balls = 4, yellow balls = 5 and pink balls = 6
Total balls = 4 + 5 + 6 = 15
Total possible outcomes from selecting of 2 balls out of 15 balls = 15C2 =
= 105
Total favourable outcomes from selecting of 2 balls out of 4 orange and 6 pink balls = 10C2
=
= 45
∴ Required probability =
= (Ans.)
Exam Aid Bank Written Math (4th Edition)
Page 108