Question 12 Marks
A bag contains 8 marbles of which 3 are blue and 5 are red. One marble is drawn at random, its cooler is noted and the marble is replaced in the bag. A marble is again drawn from the bag and its colour is noted. Find the probability that the marble will be,
Blue and red in any order.
AnswerBag contains 3 blue, 5 red marble. One marble is drawn, its colour noted and replaced then again a marble drawn and its colour is noted.
P (Blue and red in any order)
$=\text{P}\big((\text{B}\cap\text{R})\cup(\text{R}\cap\text{B})\big)$
$=\text{P}(\text{B}\cap\text{R})+\text{P}(\text{R}\cap\text{B})$
$=\text{P}(\text{B})\times\text{P}(\text{R})+\text{P}(\text{R})\times\text{P}(\text{B})$
$=\frac{3}{8}\times\frac{5}{8}+\frac{5}{8}\times\frac{3}{8}$
$=\frac{30}{64}$
$=\frac{15}{32}$
Required probability $=\frac{15}{32}$
View full question & answer→Question 22 Marks
A bag contains 8 marbles of which 3 are blue and 5 are red. One marble is drawn at random, its cooler is noted and the marble is replaced in the bag. A marble is again drawn from the bag and its colour is noted. Find the probability that the marble will be,
Blue followed by red.
AnswerBag contains 3 blue, 5 red marble. One marble is drawn, its colour noted and replaced then again a marble drawn and its colour is noted.
P (Blue followed by red)
$=\text{P}(\text{B}\cap\text{R})$
$=\text{P}(\text{B})\times\text{P}(\text{R})$
$=\frac{3}{8}\times\frac{5}{8}$
$=\frac{15}{64}$
Required probability $=\frac{15}{64}$
View full question & answer→Question 32 Marks
If $\text{P(A)}=\frac{6}{11},\text{P(B)}=\frac{5}{11}$ and $\text{P}(\text{A}\cap\text{B})=\frac{7}{11},$ find
$\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)$
AnswerGiven,
$\text{P(A)}=\frac{6}{11},\text{P(B)}=\frac{5}{11},\text{P}(\text{A}\cup\text{B})=\frac{7}{11}$
$\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(A)}}$
$=\frac{\frac{4}{11}}{\frac{6}{11}}$
$\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{2}{3}$
View full question & answer→Question 42 Marks
6 boys and 6 girls sit in a row at random. Find the probability that all the girls sit together.
AnswerHere, 6 bhoys and 6 girls can be arranged in a line in 12! ways.
Total possible outcomes = 12!
Consider 6 girls as a single element X.
Now, 6 boys and X can be arranged in aline in 7! ways and girls can be arranged in 6! ways among them.
P(all girls are together) $=\frac{7!\times6!}{12!}$
$=\frac{7\times6\times5\times4\times3\times2\times1\times6\times5\times4\times3\times2\times1}{12\times11\times10\times9\times8\times7\times6\times5\times4\times3\times2\times1}$
$=\frac{1}{11\times12}$
$=\frac{1}{132}$
View full question & answer→Question 52 Marks
If $\text{P(A)}=\frac{6}{11},\text{P(B)}=\frac{5}{11}$ and $\text{P}(\text{A}\cap\text{B})=\frac{7}{11},$ find
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$
AnswerGiven,
$\text{P(A)}=\frac{6}{11},\text{P(B)}=\frac{5}{11},\text{P}(\text{A}\cup\text{B})=\frac{7}{11}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$
$=\frac{\frac{4}{11}}{\frac{5}{11}}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{4}{5}$
View full question & answer→Question 62 Marks
Given two independent events A and B such that P(A) = 0.3 and P(B) = 0.6. Find
$\text{P}(\text{A}\cup\text{B})$
AnswerGiven that A and B are independent events and P(A) = 0.3 and P(B) = 0.6
$\text{P}(\text{A}\cup\text{B})=\text{P(A)} + \text{P(B)} - \text{P}(\text{A}\cap\text{B})$
$=0.3+0.6-0.18$
$\text{P}(\text{A}\cup\text{B})=0.72$
View full question & answer→Question 72 Marks
Write the probability that a number selected at random from the set of first 100 natural numbers is a cube.
AnswerNumber of cubes in first 100 natural numbers = 1,8,27,64
So, there are 4 cubes in first 100 natural numbers. Pgetting a cube from a set of first 100 natural numbers $=\frac{4}{100}=\frac{1}{25}$
View full question & answer→Question 82 Marks
Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that.
One of them is black and other is red.
AnswerGiven,
Box contains 10 black and 8 red balls.
Two balls are drawn with replacement.
P (one of them red and other black)
$=\text{P}\big((\text{B}\cap\text{R})\cup(\text{R}\cap\text{B})\big)$
$=\text{P}(\text{B}\cap\text{R})+\text{P}(\text{R}\cap\text{B})$
$=\text{P(B) }\text{P(R)}+\text{P(R) }\text{P(B)}$
$=\frac{10}{18}\times\frac{8}{18}+\frac{8}{18}\times\frac{10}{18}$
$=\frac{20+20}{81}$
$=\frac{40}{81}$
Required probability $=\frac{40}{81}$
View full question & answer→Question 92 Marks
A bag contains 4 white balls and 2 black balls. Another contains 3 white balls and 5 black balls. If one ball is drawn from each bag, find the probability that,
Both are black.
AnswerBag A has 4 white balls and 2 black balls;
Bag B has 3 white balls and 5 black balls;
$P (A_B$ and $B_B) = P(A_B) P(B_B)$
$=\frac{2}{6}\times\frac{5}{8}=\frac{5}{24}$
View full question & answer→Question 102 Marks
A husband and wife appear in an interview for two vacancies for the same post. The probability of husband's selection is $\frac{1}{7}$ and that of wife's selection is $\frac{1}{5}$. What is the probability that,
Only one of them will be selected?
AnswerGiven, Probability of Husband's (H) selection $=\frac{1}{7}$
$\text{P(H)}=\frac{1}{7}$
Probability of Wife's (W) selection $=\frac{1}{5}$
$\text{P(W)}=\frac{1}{5}$
P(Only one of them will be selected)
$=\text{P}\Big[(\text{H}\cap\overline{\text{W}})\cup(\overline{\text{H}}\cap\text{W})\Big]$
$=\text{P}(\text{H}\cap\overline{\text{W}})+\text{P}(\overline{\text{H}}\cap\text{W})$
$=\text{P(H)}\text{ P}(\overline{\text{W}})+\text{P}(\overline{\text{H}})\text{ P(W)}$
$=\text{P(H)}[1-\text{P(W)}]+[1-\text{P(H)}]\text{ P(W)}$
$=\frac{1}{7}\Big[1-\frac{1}{5}\Big]+\Big[1-\frac{1}{7}\Big]\frac{1}{5}$
$=\frac{1}{7}\times\frac{4}{5}+\frac{6}{7}\times\frac{1}{5}$
$=\frac{10}{35}$
$=\frac{2}{7}$
Required probability $=\frac{2}{7}$
View full question & answer→Question 112 Marks
Two cards are drawn without replacement from a pack of 52 cards. Find the probability that.
Both are kings.
AnswerIn a deck of 52 cards. there are 4 kings. Two cards are drawn without replacement,
A = First card is king
B = Second card is king
P (Both drawn cards are king)
$=\text{P}(\text{A) }\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)$
$=\frac{4}{52}\times\frac{3}{51}$
$=\frac{1}{221}$
Required Probability $=\frac{1}{221}$
View full question & answer→Question 122 Marks
If A and B are two events write the expression for the probability of occurrence of exactly one of two events.
AnswerP(exaxtly one of 2 events) $=\text{P}(\text{A}\cup\text{B})-\text{P}(\text{A}\cap\text{B})$
$=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})-\text{P}(\text{A}\cap\text{B})$
$=\text{P(A)}+\text{P(B)}-2\text{P}(\text{A}\cap\text{B})$
View full question & answer→Question 132 Marks
An unbiased die with face marked 1, 2, 3, 4, 5, 6 is rolled four times. Out of 4 face values obtained, find the probability that the minimum face value is not less than 2 and the maximum face value is not greater than 5.
AnswerDie is rolled 4 times
n(S) = Number of elements in samplw space
n(S) = 6 × 6 × 6 × 6
Given,
Number obtained on face are not less than 2 and greater than 5.
E = Obtaining 2, 3, 4, 5 on die in four throw
n(E) = 4 × 4 × 4 × 4
$\text{P(E)}=\frac{\text{n(E)}}{\text{n(S)}}$
$=\frac{4\times4\times4\times4}{6\times6\times6\times6}$
$=\frac{16}{81}$
View full question & answer→Question 142 Marks
A bag contains 4 red and 5 black balls, a second bag contains 3 red and 7 black balls. One ball is drawn at random from each bag, find the probability that the,
Balls are of the same colour.
AnswerGiven,
Bag (1) contains 4 red and 6 black balls.
Bag (2) contains 3 red and 7 black balls
One ball is drawn ar random from each bag.
P (Balls are of the same colour)
$=\text{P}\big((\text{B}_1\cap\text{B}_2)\cup(\text{R}_1\cap\text{R}_2)\big)$
$=\text{P}(\text{B}_1\cap\text{B}_2)+\text{P}(\text{R}_1\cap\text{R}_2)$
$=\text{P}(\text{B}_1)\text{P}(\text{B}_2)+\text{P}(\text{R}_1)\text{P}(\text{R}_2)$
$=\frac{5}{9}\times\frac{7}{10}+\frac{4}{9}\times\frac{3}{10}$
$=\frac{35}{90}+\frac{12}{90}$
$=\frac{47}{90}$
Required probability $=\frac{47}{90}$
View full question & answer→Question 152 Marks
If two events A and B are such that $\text{P}(\overline{\text{A}})=0.3,\text{P(B)}=0.4$ and $\text{P}(\text{A}\cap\overline{\text{B}})=0.5$ find $\text{P}\Big(\frac{\text{B}}{\overline{\text{A}}\cap\overline{\text{B}}}\Big).$
AnswerAccording to Baye's Theorem
$\text{P}\Big(\frac{\text{B}}{\overline{\text{A}}\cap\overline{\text{B}}}\Big)=\frac{\text{P}(\text{B}\cap(\overline{\text{A}}\cap\overline{\text{B}}))}{\text{P}(\overline{\text{A}}\cap\overline{\text{B}})}$
$=\frac{\text{P}(\text{B}(\overline{\text{A}\cap\text{B}}))}{\text{P}(\overline{\text{A}}\cap\overline{\text{B}})}$
$=\frac{\text{P}(\overline{\text{B}}(\overline{\text{A}\cap\text{B}}))}{\text{P}(\overline{\text{A}}\cap\overline{\text{B}})}$
$=\frac{\text{P}(\overline{\text{B}}\cup(\text{A}\cup\text{B}))}{\text{P}(\overline{\text{A}}\cap\overline{\text{B}})}$
Now $\overline{\text{B}}\cap\text{B}=\cup=\phi$
So, $\text{P}(\overline{\text{B}}\cup(\text{A}\cup\text{B}))=\phi$
$\therefore\ \text{P}\Big(\frac{\text{B}}{\overline{\text{A}}\cap\overline{\text{B}}}\Big)=0$
View full question & answer→Question 162 Marks
When three dice are thrown, write the probability of getting 4 or 5 on each of the dice simultaneously.
AnswerThere dice are thrown
Given, 4 or 5 on each of the dice simultaneously
= {(4, 4, 4), (4, 4, 5), (4, 5, 4), (4, 5, 5), (5, 4, 4), (5, 4, 5), (5, 5, 4), (5, 5, 5)}
n(E) = 8
n(S) = 216
$\text{P(E)}=\frac{8}{216}$
$=\frac{1}{27}$
Required probability $=\frac{1}{27}$
View full question & answer→Question 172 Marks
Given two independent events A and B such that P(A) = 0.3 and P(B) = 0.6. Find
$\text{P}(\text{A}\cap\text{B})$
AnswerGiven that A and B are independent events and P(A) = 0.3 and P(B) = 0.6
$\text{P}(\text{A}\cap\text{B})=\text{P(A) }\text{P(B)}$
[Since, A and B are indenpendent events]
$=0.3\times0.6$
$\text{P}(\text{A}\cap\text{B})=0.18$
View full question & answer→Question 182 Marks
If A and B are two independent events such that P(A) = 0.3 and $=0.8\text{P}(\text{A}\cap\overline{\text{B}})$ Find P(B).
AnswerA and B are two independent events
$\therefore\ \text{P}(\text{A}\cap\overline{\text{B}})=\text{P(A)}+\text{P}(\overline{\text{B}})-\text{P}(\text{A}\cap\overline{\text{B}})$
$\Rightarrow\ 0.8=0.3+[1-\text{P(B)}]-\text{P(A)}\text{P}(\overline{\text{B}})$
$\Rightarrow 0.5=1-\text{P(B)}-0.3[1-\text{P(B)}]$
$\Rightarrow 0.5=1-\text{P(B)}=0.3+0.3\text{P(B)}$
$\Rightarrow\ 0.5= 0.7-\text{P(B)}[1-0.3]$
$\Rightarrow\ 0.7\text{P(B)} = 0.2$
$\Rightarrow\ \text{P(B)}=\frac{0.2}{0.7}=\frac{2}{7}$
View full question & answer→Question 192 Marks
Fatima and John appear in an interview for two vacancies for the same post. The probability of Fatima's selection is $\frac{1}{7}$ and that of John's selection is $\frac{1}{5}$. What is the probability that,
Both of them will be selected?
AnswerGiven,
Probability of Fatima's (F) selection $=\frac{1}{7}$
$\text{P(F)}=\frac{1}{7}\Rightarrow\ \text{P}(\overline{\text{F}})=\frac{6}{7}$
Probability of John's (J) selection $=\frac{1}{5}$
$\text{P(F)}=\frac{1}{5}\Rightarrow\ \text{P}(\overline{\text{F}})=\frac{4}{5}$
P(Boht of them selected)
$=\text{P}(\text{F}\cap\text{J})$
$=\text{P}(\text{F})\text{P}(\text{J})$
$=\frac{1}{7}\times\frac{1}{5}$
$=\frac{1}{35}$
Required probability $=\frac{1}{35}$
View full question & answer→Question 202 Marks
Two cards are drawn without replacement from a pack of 52 cards. Find the probability that.
The first is a heart and second is red.
AnswerThere are 13 heart and 26 red cards
Hearts are also red.
A = first card is heart.
B = second card is red.
P (First card is heart and second is red)
$=\text{P(A) }\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)$
$=\frac{13}{52}\times\frac{25}{51}$
$=\frac{25}{204}$
Required probability $=\frac{25}{204}$
View full question & answer→Question 212 Marks
Let A and B be two independent events such that $P(A) = p_1$ and $P(B) = p_2.$ Describe in words the events whose probabilities are:
$(1 - p_1)p_2.$
AnswerAs, $(1 - p_1)p_2$
$=[1-\text{P(A)}]\times\text{P(B)}=\text{P}(\overline{\text{A}})\times\text{P(B)}$
And, A and B are independent events.
i.e., $\text{P}(\overline{\text{A}})\times\text{P}(\text{B})=\text{P}(\text{A}\cap\text{B})$
So, $\text{P}(\text{A}\cap\text{B})=\text{p}_1\text{p}_2$
Hence, $(1 - p_1)p_2 = P (A$ does not occur, but $B$ occurs$).$
View full question & answer→Question 222 Marks
Given two independent events A and B such that P(A) = 0.3 and P(B) = 0.6. Find
$\text{P}(\text{A}\cap\overline{\text{B}})$
AnswerGiven that A and B are independent events and P(A) = 0.3 and P(B) = 0.6
$\text{P}(\text{A}\cap\overline{\text{B}})=\text{P(A)}-\text{P}(\text{A}\cap\text{B})$
$=0.3 - 0.18$
$\text{P}(\text{A}\cap\overline{\text{B}})=0.12$
View full question & answer→Question 232 Marks
An urn contains 7 red and 4 blue balls. Two balls are drawn at random with replacement. Find the probability of getting,
2 red balls.
AnswerAn urn contains 7 red and 4 blue balls.
Two balls are drawn with replacement.
P(Getting 2 red balls)
$=\text{P}(\text{R}_1)\times\text{P}(\text{R}_2)$
$=\frac{7}{11}\times\frac{7}{11}$
$=\frac{49}{121}$
Required probability $=\frac{49}{121}$
View full question & answer→Question 242 Marks
A husband and wife appear in an interview for two vacancies for the same post. The probability of husband's selection is $\frac{1}{7}$ and that of wife's selection is $\frac{1}{5}$. What is the probability that,
None of them will be selected?
AnswerGiven, Probability of Husband's (H) selection $=\frac{1}{7}$
$\text{P(H)}=\frac{1}{7}$
Probability of Wife's (W) selection $=\frac{1}{5}$
$\text{P(W)}=\frac{1}{5}$
P(None of them selected)
$=(\overline{\text{H}}\cap\overline{\text{W}})$
$=\text{P}(\overline{\text{H}})\text{ P}(\overline{\text{W}})$
$=(1-\text{P(H)})(1-\text{P(W)})$
$=\Big(1-\frac{1}{7}\Big)\Big(1-\frac{1}{5}\Big)$
$=\frac{6}{7}\times\frac{4}{5}$
$=\frac{24}{35}$
Required probability $=\frac{24}{35}$
View full question & answer→Question 252 Marks
Out of 100 students, two sections of 40 and 60 are formed. If you and your friend are among 100 students, what is the probability that:
You both enter the different sections?
AnswerP (Both enter different section)
= 1 - P(Both enter same section)
$=1-\frac{17}{33}$
$=\frac{16}{33}$
P(Both eneter different section) $=\frac{16}{33}$
View full question & answer→Question 262 Marks
A bag contains 8 marbles of which 3 are blue and 5 are red. One marble is drawn at random, its cooler is noted and the marble is replaced in the bag. A marble is again drawn from the bag and its colour is noted. Find the probability that the marble will be,
Of the same colour.
AnswerBag contains 3 blue, 5 red marble. One marble is drawn, its colour noted and replaced then again a marble drawn and its colour is noted.
P (Of the same colour)
$=\text{P}\big((\text{R}_1\cap\text{R}_2)\cup(\text{B}_1\cap\text{B}_2)\big)$
$=\text{P}(\text{R}_1)\times\text{P}(\text{R}_2)+\text{P}(\text{B}_1)\times\text{P}(\text{B}_2)$
$=\frac{5}{8}\times\frac{5}{8}+\frac{3}{8}\times\frac{3}{8}$
$=\frac{25+9}{64}$
$=\frac{34}{64}$
$=\frac{17}{32}$
View full question & answer→Question 272 Marks
An urn contains 7 red and 4 blue balls. Two balls are drawn at random with replacement. Find the probability of getting,
One red and one blue ball.
AnswerAn urn contains 7 red and 4 blue balls.
Two balls are drawn with replacement.
P(Getting 2 blue balls)
$=\text{P}\big((\text{R}\cap\text{B})\cup(\text{B}\cap\text{R})\big)$
$=\text{P(R)}\times\text{P(B)}+\text{P(B)}\times\text{P(R)}$
$=\frac{7}{11}\times\frac{4}{11}+\frac{4}{11}\times\frac{7}{11}$
$=\frac{28+28}{121}$
$=\frac{56}{121}$
Required probability $=\frac{56}{121}$
View full question & answer→Question 282 Marks
If $\text{P(A)}=0.3,\text{P(B)}=0.6,\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=0.5,$ find $\text{P}(\text{A}\cup\text{B}).$
AnswerGiven,
$\text{P(A)}=0.3,\text{P(B)}=0.6,\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=0.5,$
$\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(A)}}$
$0.5=\frac{\text{P}(\text{A}\cap\text{B})}{0.3}$
$\text{P}(\text{A}\cap\text{B})=0.5\times0.3$
$\text{P}(\text{A}\cap\text{B})=0.15$
$\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$=0.3+0.6+0.15$
$=0.75$
$\text{P}(\text{A}\cup\text{B})=0.75$
View full question & answer→Question 292 Marks
Three digit numbers are formed with the digits 0, 2, 4, 6 and 8. Write the probability of forming a three digit number with the same digits.
AnswerTotal 3-digit numbers that can be made out of 0, 2, 4, 5 and 8 = 4 × 5 × 5 (hundreds place cannot be filled with 0)
= 100
But 222, 444, 666 and 888 are four numbers, which have the same digits at all places.
P(3-digit number having same digits at all places) $=\frac{4}{100}$
$=\frac{1}{25}$
View full question & answer→Question 302 Marks
An urn contains 7 red and 4 blue balls. Two balls are drawn at random with replacement. Find the probability of getting,
2 Blue balls.
AnswerAn urn contains 7 red and 4 blue balls.
Two balls are drawn with replacement.
P(Getting 2 blue balls)
$=\text{P}(\text{B}_1\cap\text{B}_2)$
$=\text{P}(\text{B}_1)\times\text{P}(\text{B}_2)$
$=\frac{4}{11}\times\frac{4}{11}$
$=\frac{16}{121}$
Required probability $=\frac{16}{121}$
View full question & answer→Question 312 Marks
Kamal and Monica appeared for an interview for two vacancies. The probability of Kamal's selection is $\frac{1}{3}$ and that of Monika's selection is $\frac{1}{5}$. Find the probability that.
Both of them will be selected.
AnswerP(Kamal gets selected) $=\text{P(A)}=\frac{1}{3}$
P(Monica gets selected) $=\text{P(B)}=\frac{1}{5}$
P(Both get selected) = P(A) × P(B)
$=\frac{1}{3}\times\frac{1}{5}$
$=\frac{1}{15}$
View full question & answer→Question 322 Marks
Given two independent events A and B such that P(A) = 0.3 and P(B) = 0.6. Find
$\text{P}(\overline{\text{A}}\cap\text{B})$
AnswerGiven that A and B are independent events and P(A) = 0.3 and P(B) = 0.6
$\text{P}(\overline{\text{A}}\cap\text{B})=\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$=0.6 - 0.18$
$\text{P}(\overline{\text{A}}\cap\text{B})=0.42$
View full question & answer→Question 332 Marks
A bag contains 4 white balls and 2 black balls. Another contains 3 white balls and 5 black balls. If one ball is drawn from each bag, find the probability that,
One is while and one is black.
AnswerBag A has 4 white balls and 2 black balls;
Bag B has 3 white balls and 5 black balls;
$P (A_W$ and $B_B$ or $A_B$ and $B_W)$
$= P(A_W) P(B_B) + P(A_B) P(B_W)$
$=\frac{4}{6}\times\frac{5}{8}+\frac{2}{6}\times\frac{3}{8}$
$=\frac{20}{48}+\frac{6}{48}$
$=\frac{26}{48}=\frac{13}{24}$
View full question & answer→Question 342 Marks
A bag contains 6 black and 3 white balls. Another bag contains 5 black and 4 white balls. If one ball is drawn from each bag, find the probability that these two balls are of the same colour.
AnswerGiven:
Bag 1 = (3W + 6B) balls
Bag 2 = (5B + 4W) balls
P(balls of same colour are drawn) = P(both black) + P(both white)
$=\frac{6}{9}\times\frac{5}{9}+\frac{3}{9}\times\frac{4}{9}$
$=\frac{30}{81}+\frac{12}{81}$
$=\frac{42}{81}$
$=\frac{14}{27}$
View full question & answer→Question 352 Marks
Three cards are drawn with replacement from a well shuffled pack of cards. Find the probability that the cards drawn are king, queen and jack.
Answer$\text{P (king)}=\frac{4}{52}$
$\text{P (queen)}=\frac{4}{52}$
$\text{P (jack)}=\frac{4}{52}$
These cards can be drawn in $^3P_3$ ways.
$\text{P (king, queen, and jack)}=\frac{4}{52}\times\frac{4}{52}\times\frac{4}{52}\times {^{3}}\text{P}_3$
$=\frac{3!}{2197}$
$=\frac{6}{2197}$
View full question & answer→Question 362 Marks
A four digit number is formed using the digits 1, 2, 3, 5 with no repetitions. Write the probability that the number is divisible by 5.
AnswerTo be divisibel by 5 ones place sholud be 5
There are 3 place remaining which can be filled in 3! = 6 ways
So, 6 number can be formed out of 1, 2, 3 and 5, which are divisible by 5.
Total 4 - digit numbers = 4! = 24
P (4-digit number divisible by 5) $=\frac{6}{24}$
$=\frac{1}{4}$
View full question & answer→Question 372 Marks
A bag contains 4 white balls and 2 black balls. Another contains 3 white balls and 5 black balls. If one ball is drawn from each bag, find the probability that,
Both are white.
AnswerBag A has 4 white balls and 2 black balls;
Bag B has 3 white balls and 5 black balls;
$P (A_W$ and $B_W) = P(A_W) P(B_W)$
$=\frac{4}{6}\times\frac{3}{8}=\frac{1}{4}$
View full question & answer→Question 382 Marks
Two cards are drawn from a well shuffled pack of 52 cards, one after another without replacement. Find the probability that one of these is red card and the other a black card?
AnswerP(One red and one black) = P(first red and second black) + P(first black and second red)
$=\frac{26}{5}\times\frac{26}{51}+\frac{26}{52}\times\frac{26}{51}$
[Without replacement]
$=\frac{13}{51}+\frac{13}{51}$
$=\frac{26}{51}$
View full question & answer→Question 392 Marks
A card is drawn from a well-shulffled deck of 52 cards. The outcome is noted, the card is replaced and the deck reshuffled. Another card is then drawn from the deck.
What is the probability that both the cards are of the same suit?
AnswerP(both the cards are of same suit) = P(both the cards are of diamond) + P(both the cards are of spade) + P(both the cards are of club) + P(both the cards are of hear)
$=\frac{13}{52}\times\frac{13}{52}+\frac{13}{52}\times\frac{13}{52}+\frac{13}{52}\times\frac{13}{52}+\frac{13}{52}\times\frac{13}{52}$
$=\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}$
$=\frac{4}{16}$
$=\frac{1}{4}$
View full question & answer→Question 402 Marks
If $\text{P(A)}=\frac{7}{13},\text{P(B)}=\frac{9}{13}$ and $\text{P}(\text{A}\cap\text{B})=\frac{4}{13},$ find $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big).$
AnswerGiven:
$\text{P(A)}=\frac{7}{13}$
$\text{P(B)}=\frac{9}{13}$
$\text{P}(\text{A}\cap\text{B})=\frac{4}{13}$
Now,
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$
$\Rightarrow\ \text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\frac{4}{13}}{\frac{9}{13}}=\frac{4}{9}$
View full question & answer→Question 412 Marks
In a competition A, B and C are participating. The probability that A wins is twice that of B, the probability that B wins is twice that of C. Find the probability that A losses.
AnswerLet P(A wins) = x
So, P(B wins) = 2x
P(A wins) = 2P(B wins)
= 2(2x)
P(A wins) = 4X
P(A wins) + P(B wins) + P(C wins) = 1
⇒ 4x + 2x+ x = 1
⇒ 7x = 1
$\Rightarrow\ \text{x}=\frac{1}{7}$
$\text{P(A wins)}=4\text{x}$
$=\frac{4}{7}$
$\text{P(A losses)}=1-\text{P(A wins)}$
$=1-\frac{4}{7}$
$=\frac{3}{7}$
Required probability $=\frac{3}{7}$
View full question & answer→Question 422 Marks
The probability of student A passing an examination is $\frac{2}{9}$ and of student B passing is $\frac{5}{9}$. Assuming the two events: 'A passes', 'B passes' as independent, find the probability of:
Only A passing the examination.
AnswerGiven,
The probability of A passing exam $=\frac{2}{9}$
The probability of B passing exam $=\frac{5}{9}$
$\Rightarrow\ \text{P(A)}=\frac{2}{9},\text{P(B)}=\frac{5}{9}$
P (Only A passing the exam)
$=\text{P}(\text{A}\cap\overline{\text{B}})$
$=\text{P(A)}\text{P(B)}$
$=\text{P(A)}[1-\text{P(B)}]$
$=\frac{2}{9}\Big(1-\frac{5}{9}\Big)$
$=\frac{2}{9}\Big(\frac{4}{9}\Big)$
$=\frac{8}{81}$
View full question & answer→Question 432 Marks
A ordinary cube has four plane faces, one face marked 2 and another face marked 3, find the probability of getting a total of 7 in 5 throws.
AnswerA cube has total 6 faces.
Total possible outcomes in 5 throws $= 6 \times 6\times 6 \times 6 \times 6 = (6)^5$
The only way of getting 7 is by getting two 2s and one 3.
Total possible ways $=\frac{\text{P}^5_3}{2!}$
$=\frac{5\times4\times3\times2\times1}{2\times1\times2\times1}$
$=30$
Now,
$P($getting $7$ in $5$ throws$)=\frac{30}{6^5}=\frac{5}{6^4}$
View full question & answer→Question 442 Marks
Kamal and Monica appeared for an interview for two vacancies. The probability of Kamal's selection is $\frac{1}{3}$ and that of Monika's selection is $\frac{1}{5}$. Find the probability that.
Only one of them will be selected.
AnswerP (Kamal gets selected) $=\text{P(A)}=\frac{1}{3}$
P (Monica gets selected) $=\text{P(B)}=\frac{1}{5}$
P (One of them gets selected) $=\text{P}(\overline{\text{A}})\text{ P(B)}+\text{P}(\overline{\text{B}})\text{ P(A)}$
$=\text{P(B)}[1-\text{P(A)}]+\text{P(A)}[1-\text{P(B)}]$
$=\frac{1}{5}\Big(1-\frac{1}{3}\Big)+\frac{1}{3}\Big(1-\frac{1}{5}\Big)$
$=\frac{2}{15}+\frac{4}{15}$
$=\frac{6}{15}=\frac{2}{5}$
View full question & answer→Question 452 Marks
Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that.
First ball is black and second is red.
AnswerGiven,
Box contains 10 black and 8 red balls.
Two balls are drawn with replacement.
P (First ball is black and second is red)
$=\text{P}(\text{B}\cap\text{R})$
$=\text{P(B)}\text{ P(R)}$
$=\frac{10}{18}\times\frac{8}{18}$
$=\frac{20}{81}$
Required probability $=\frac{20}{81}$
View full question & answer→Question 462 Marks
Kamal and Monica appeared for an interview for two vacancies. The probability of Kamal's selection is $\frac{1}{3}$ and that of Monika's selection is $\frac{1}{5}$. Find the probability that.
none of them will be selected.
AnswerP(Kamal gets selected) $=\text{P(A)}=\frac{1}{3}$
P(Monica gets selected) $=\text{P(B)}=\frac{1}{5}$
P(none of them get selected) $=\text{P}(\overline{\text{A}})\times\text{P}(\overline{\text{B}})$
$=[1-\text{P(A)}][1-\text{P(B)}]$
$=\Big(1-\frac{1}{3}\Big)\Big(1-\frac{1}{5}\Big)$
$=\frac{2}{3}\times\frac{4}{5}$
$=\frac{8}{15}$
View full question & answer→Question 472 Marks
Let A and B be two independent events such that $P(A) = p_1$ and $P(B) = p_2$. Describe in words the events whose probabilities are:
$p_1p_{2}.$
AnswerAs, $p_1p_2 = P(A) \times P(B)$
And, A and B are independent events.
i.e., $\text{P(A)}\times\text{P(B)}=\text{P}(\text{A}\cap\text{B})$
So, $\text{P}(\text{A}\cap\text{B})=\text{P}_1\text{P}_2$
Hence, $p_1p_2 = P(A$ and $B$ occur$).$
View full question & answer→Question 482 Marks
Assume that each child born is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that,
- The youngest is a girl.
- At least one is a girl?
AnswerLet a and g rapresent the boyy and the girl child respectively. if a family has two children, the samplw space will be
S = {(b, b), (b, g), (g, b), (g, g)}
Let a be the event that both children are girls.
$\therefore\ \text{A}=\big\{(\text{g},\text{g})\big\}$
- Let B be the event that the youngest child is a girl.
$\therefore\text{B}=\big[(\text{b},\text{g}),(\text{g},\text{g})\big]$
$\Rightarrow\ \text{A}\cap\text{B}=\big\{(\text{g},\text{g})\big\}$
$\therefore\text{P(B)}=\frac{2}{4}=\frac{1}{2}$
$\text{P}(\text{A}\cap\text{B})=\frac{1}{4}$
The conditional probability that both are girls, given that the youngest child is a girl, is given by p (A | B).
$\text{P}(\text{A}|\text{B})=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{2}$
Therefore, the required probability is $\frac{1}{2}$.
- Let C be the event that at least one child is a girl.
$\therefore\text{C}=\big\{(\text{b},\text{g}),(\text{g},\text{b}),(\text{g},\text{g})\big\}$
$\Rightarrow\text{A}\cap\text{C}=\big\{\text{g},\text{g}\big\}$
$\Rightarrow\text{P(C)}=\frac{3}{4}$
$\text{P}(\text{A}\cap\text{C})=\frac{1}{4}$
The conditional probability that both are girls, given that at least one child a girl, is given by P(A|C).
$\therefore \text{P}(\text{A}|\text{C})=\frac{\text{P}(\text{A}\cap\text{C})}{\text{P}(\text{C})}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}$ View full question & answer→Question 492 Marks
X is taking up subjects - Mathematics, Physics and Chemistry in the examination. His probabilities of getting grade A in these subjects are 0.2, 0.3 and 0.5 respectively. Find the probability that he gets,
Grade A in no subject.
AnswerGiven,
Probability of getting A grade in mathematics (m) = 0.2
⇒ P(m) = 0.2
Probability of getting A grade in physics (p) = 0.3
⇒ P(p) = 0.3
Probability of getting A grade in chemistry (P) = 0.5
⇒ P(C) = 0.5
P(Getting A in no subject)
$=\text{P}(\overline{\text{m}}\cap\overline{\text{p}}\cap\overline{\text{c}}\big)$
$=\text{P}(\overline{\text{m}})+\text{P}(\overline{\text{p}})+\text{P}(\overline{\text{c}}\big)$
$=\big(1-\text{P(m)}\big)\big(1-\text{P(p)}\big)\big(1-\text{P(c)}\big)$
$=(1-0.2)(1-0.3)(1-0.5)$
$=(0.8)(0.7)(0.5)$
$=0.28$
Required probability = 0.28
View full question & answer→Question 502 Marks
Given two independent events A and B such that P(A) = 0.3 and P(B) = 0.6. Find
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$
AnswerGiven that A and B are independent events and P(A) = 0.3 and P(B) = 0.6
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$
$=\frac{0.18}{0.6}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=0.3$
View full question & answer→