Download CBSE Sample Paper 2024-25 for class 12th to 8th

Probability Test - 55

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Probability Test - 55

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Weekly Quiz Competition

Question 1
1 / -0

If the integers $$'m'$$ and $$'n'$$ are chosen at random from $$1$$ to $$100$$ then the probability that $$7^{m}+7^{n}$$ is divisible by $$5$$ is ?

A
$$1/5$$

B
$$1/7$$

C
$$1/4$$

D
$$1/49$$

Solution

$${ 7 }^{ m }$$ can be 7,49,343,2401,16807.....
Last digits are
$${ 7 }^{ 4m }$$=1, $${ 7 }^{ 4m+1 }$$=7,$${ 7 }^{ 4m+2 }$$=9,$${ 7 }^{ 4m+3 }$$=3
Therefore ,$${ 7 }^{ m }$$ has last digits 7,9,3,1
For $${ 7 }^{ m }$$+$${ 7 }^{ n }$$ to be divisible by 5 last digit must be combination of 7,3 and 9,1.
Total possible way to choose m and n =100*100
There are 25 types of 4m between 1 to 10025 types of 4m+1 between 1 to 100,
25 types of 4m+2 between 1 to 100,25 types of 4m+3 between 1 to 100
So ways to choose m and n such that last digits are 7,3 or 3,7=2*25*25
So ways to choose m and n such that last digits are 1,9 or 9,1=2*25*25
Total ways =4*25*25
Probability=4*25*25/(100*100)=1/4
Question 2
1 / -0

Consider 5 independent bernoulli's trials each with probability p of success. If the probability of one failure is greater than or equal to $${{31} \over {32}}$$, then p lies in the interval

A
$$\left[ {{{11} \over {12}},1} \right]$$

B
$$\left[ {{1 \over 2},{3 \over 4}} \right]$$

C
$$\left[ {{3 \over 4},{{11} \over {12}}} \right]$$

D
$$\left[ {0,{1 \over 2}} \right]$$

Solution
Question 3
1 / -0

A fair dice is rolled three times, then the probability of getting a number larger than the previous number is

A
$$\frac { 5 }{ 216 } $$

B
$$\frac { 5 }{ 54 } $$

C
$$\frac { 1 }{ 6 } $$

D
$$\frac { 5 }{ 36 } $$

Solution
Question 4
1 / -0

Two non-negative integers are chosen at random from the set of non-negative integers with replacement. The probability that the sum of their squares is divisible by $$10$$ is

A
$$7/100$$

B
$$6/50$$

C
$$9/50$$

D
$$12/50$$

Solution

Hello student,
Please find the answer to your question below
Given Two non negative integers are chosen at random.
Let the 2 non negative integers be a,b then 0<=a,b<=9
Number of ways =10*10=100
The probability that their sum of their square is divisible by 10 is=(1,3),(2,6),(7,1),(4,8),(3,9),(2,4),(5,5) are 7
Probability=7/100
Question 5
1 / -0

Directions For Questions

A player tosses a coin and scores one point for every head and two points for every tail that turns up. He plays on until his score reaches or passes $$n.{P}_{n}$$ denotes the probability of gegging a score of exactly $$n$$.

...view full instructions

The value of $${P}_{n}$$ is equal to

A
$$(1/2)[{P}_{n-1}+{P}_{n-2}]$$

B
$$(1/2)[2{P}_{n-1}+{P}_{n-2}]$$

C
$$(1/2)[{P}_{n-1}-{P}_{n-2}]$$

D
$$none\ of\ these$$

Solution

We can score exactly n in two ways: by scoring exactly n−1, and then throwing a head; or by not scoring n-1.
Let Pn be the probability that our random walk passes through
n. Since the first step is 1 or 2, each with probabiliyt 1/2
So ,$${ P }_{ n+1 }$$=1/2$${ P }_{ n }$$+1/2$${ P }_{ n-1 }$$
With inital condition $${ P }_{ 0 }$$=1,$${ P }_{ 1 }$$=1/2
Question 6
1 / -0

If the integers $$m$$ and $$n$$ are chosen at random from $$1$$ to $$100$$ then the probability that $${7}^{m}+{7}^{n}$$ is divisible by $$5$$ is ?

A
$$1/5$$

B
$$1/7$$

C
$$1/4$$

D
$$1/49$$

Solution

$${ 7 }^{ m }$$ can be 7,49,343,2401,16807.....
Last digits are
$${ 7 }^{ 4m }$$=1, $${ 7 }^{ 4m+1 }$$=7,$${ 7 }^{ 4m+2 }$$=9,$${ 7 }^{ 4m+3 }$$=3
Therefore ,$${ 7 }^{ m }$$ has last digits 7,9,3,1
For $${ 7 }^{ m }$$+$${ 7 }^{ n }$$ to be divisible by 5 last digit must be combination of 7,3 and 9,1.
Total possible way to choose m and n =100*100
There are 25 types of 4m between 1 to 10025 types of 4m+1 between 1 to 100,
25 types of 4m+2 between 1 to 100,25 types of 4m+3 between 1 to 100
So ways to choose m and n such that last digits are 7,3 or 3,7=2*25*25
So ways to choose m and n such that last digits are 1,9 or 9,1=2*25*25
Total ways =4*25*25
Probability=4*25*25/(100*100)=1/4
Question 7
1 / -0

Directions For Questions

A player tosses a coin and scores one point for every head and two points for every tail that turns up. He plays on until his score reaches or passes $$n.{P}_{n}$$ denotes the probability of gegging a score of exactly $$n$$.

...view full instructions

Which of the following is not true?

A
$${P}_{100}>2/3$$

B
$${P}_{101}<2/3$$

C
$${P}_{100}{P}_{101}>2/3$$

D
$$none\ of\ these$$

Solution

We can score exactly n in two ways: by scoring exactly n−1, and then throwing a head; or by not scoring n-1.
Let Pn be the probability that our random walk passes through
n. Since the first step is 1 or 2, each with probabiliyt 1/2
So ,$${ P }_{ n+1 }$$=1/2$${ P }_{ n }$$+1/2$${ P }_{ n-1 }$$
With inital condition $${ P }_{ 0 }$$=1,$${ P }_{ 1 }$$=1/2
We get $${ P }_{ 2 }$$=3/4 and $${ P }_{ 3 }$$=5/8
On solving we get recurrence relation as $${ P }_{ n }=\frac { 2 }{ 3 } +\frac { 1 }{ 3 } { (-1/2) }^{ n }$$
For every even number n ,$${ P }_{ n }$$>2/3
So we get, $${ P }_{ 100 }$$>2/3
Question 8
1 / -0

A boy has a collection of blue and green marbles. The number of blue marbles belong to the sets $${2,3,4,13}$$. If two marbles are chosen simultaneously and at random from his collection, then the probability that they have different colour is $$1/2$$. Possible number of blue marbles is:

A
$$2$$

B
$$3$$

C
$$6$$

D
$$10$$

Solution

Suppose, there are m blue and n green marbles.
There are (m+n)*(m+n−1)/2 ways to choose 2 marbles.
There are m*n ways to choose 2 marbles with different colours.
The probability of getting two marbles with different colours is therefore
$2 m n ( m + n ) ( m + n - 1 )$
So, the probability is 12, if and only if
$(m + n) (m + n - 1) = 4 m n$
1) m=2 : n2+3n+2=8n has no solution in N
2) m=3 : n2+5n+6=12n has the solutions 1 and 6.
3) m=6 : n2+11n+30=24n has the solutions 3 and 10.
4) m=10 : n2+19n+90=40n has the solutions 6 and 15.
Question 9
1 / -0

Two numbers x and y are chosen at random without replacement from the first $$30$$ natural numbers. The probability that $$x^2-y^2$$ is divisible by $$3$$ is?

A
$$\dfrac{3}{29}$$

B
$$\dfrac{3}{55}$$

C
$$\dfrac{5}{29}$$

D
$$\dfrac{47}{87}$$

Solution
Question 10
1 / -0

Let $$A$$ and $$B$$ be events such that $$P(\overline{A})=\dfrac{4}{5}, P(B) = \dfrac{1}{3}, P\left(\dfrac{A}{B}\right) = \dfrac{1}{6}$$, then:

A
$$P(A \cap B)$$

B
$$P(A \cup B)$$

C
$$P\left(\dfrac{B}{A}\right)$$

D
Are $$A$$ and $$B$$ independent

Solution

$$P(\bar{A})=\dfrac{4}{5}$$ then $$P(A)=\dfrac{1}{5}$$

$$P(B)=\dfrac{1}{3}$$ and $$P\left ( \dfrac{A}{B} \right ) =\dfrac{1}{6}$$

we know that $$P\left ( \dfrac{A}{B} \right )=\dfrac{P(A\cap B)}{P(B)}$$

$$\dfrac{1}{6}=\dfrac{A(A\cap B)}{1}\cdot 3$$

$$P(A\cap B)=\dfrac{1}{1^{8}}$$

we know that $$P(A)+P(B)=P(A\cup B)+P(A\cap B)$$

$$\dfrac{1}{5}+\dfrac{1}{3}-\dfrac{1}{18}=P(A\cup B))$$

$$P(A\cup B)=\dfrac{18-5+30}{90}=\dfrac{43}{90}$$

$$P\left ( \dfrac{B}{A} \right )=\dfrac{P(B\cap A)}{P(A)}=\dfrac{1\cdot 5}{18\cdot 1}=\dfrac{5}{18}$$