(A) `\frac{5}{11}`
(B)
(C)
(D)
Answer. (A)
Solution.
Let the three numbers are a, b, c and as per the question,
a + b + c = 10
As per the given condition let’s consider that ‘a’ will always be odd.
Now we will calculate all the possible solutions, i.e.; total number of
events. For this, let’s first take a = 0, then the possible values of b and c
will be,
|
b |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
|
c |
10 |
9 |
8 |
7 |
6 |
5 |
4 |
3 |
2 |
1 |
0 |
Similarly, when a = 1, then
|
b |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
|
c |
9 |
8 |
7 |
6 |
5 |
4 |
3 |
2 |
1 |
0 |
We can notice that when a=0 then we had 11 events, when a=1 then we had
10 events. So when the value of a will increase then the number of events will
decrease and there will be only one event at the end. So total number of events
will be,
t(E) = 11+10+9+8+7+6+5+4+3+2+1 = 66
But our favorable events will be the events that favor odd values of
‘a’. Here we can notice that for each odd value of ‘a’ the number of events are
even. Hence, all favorable events will be,
f(E) = 10+8+6+4+2 = 30
Hence, the required probability is,
P = `\frac{30}{66}` = `\frac{5}{11}`
Question 32.) There are five members M1, M2, M3, M4 and M5 of a library and five different books having numbers B1, B2, B3, B4, and B5 are allotted to each of the member in such a manner that M1 gets the book B1, M2 gets the book B2 and so on. At the end of the month they all are asked to submit the books allotted to them. These books will be re-allotted to them in the next month. Then find the probability that, when books are re-allotted then only M1 gets the book B1and NONE of the remaining members get the book of same number previously allotted to them.
(A) `\frac{7}{40}`
(B) `\frac{1+9}{8}`
(C) `\frac{3}{40}`
(D) `\frac{9}{40}`
Answer. (C)
Solution.
Since, there are 5 members and 5 books. So, total number of events will be = 5! = 120
As per the condition given in the question only M1 gets the previously
allotted book B1 and no other gets his previously allotted book. So, number of
favorable events will be,
= 4! (1 - `\frac{1}{1!}` + `\frac{1}{2!}` - `\frac{1}{3!}` + `\frac{1}{4!}`) = 24 (1 - 1 + `\frac{1}{2}` - `\frac{1}{6}` + `\frac{1}{24}`) = 24 (`\frac{12-4+1}{24}`) = 9
Hence the required probability = `\frac{9}{120}` = `\frac{3}{40}`
Question 33.) There are three bags B1, B2 and B3. The bag B1 contains 5 red and some green balls, B2 contains 3 red and 5green balls and B3 contains 5 red and 3 green balls. Bags B1, B2 and B3 have probabilities `\frac{3}{10}`, `\frac{3}{10}`, and `\frac{4}{10}` respectively of being chosen. A bag is selected at random and a ball is chosen at random from the bag. If the probability that the chosen ball is green is equal to `\frac{39}{80}` then find the number of green balls in bag B1.
(A) 5
(B) 4
(C) 7
(D) 3
Answer. (A)
Solution.
Let the number of green balls in box B1 is x
Given that the probabilities of choosing the bags at random are,
P(B1) =
P(B2) =
P(B3) =
The probability of choosing a green ball from bag B1 will be,
P(`\frac{G}{B1}`) = `\frac{x}{x+5}`
Similarly, the probabilities of choosing a green ball from bags B2 and
B3 will be,
P(`\frac{G}{B2}`) = `\frac{5}{8}`
P(`\frac{G}{B3}`) = `\frac{3}{8}`
Now the required probability is,
P(G) = P(B1) × P(`\frac{G}{B1}`) + P(B2) c P(`\frac{G}{B2}`)
On solving it we will get x
Hence, the number of green balls in box B1 = 5
Question 34.) Two fair dices are rolled together and the sum of the numbers on the faces is observed. Find the probability that their sum is a not a perfect square.
(A) `\frac{7}{36}`
(B) `\frac{29}{36}`
(C) `\frac{11}{36}`
(D) `\frac{2}{3}`
Answer. (B)
Solution.
When two dices are rolled then the total number of events will be = 6 × 6 = 36
Let P is the probability of getting a sum which is not a perfect square
and P̅ is the probability of getting a sum which is a perfect square. Then,
P = 1 – P̅
Now when two dices are rolled then their sum cannot be more than 12 and
there are 3 perfect squares under 12; 1, 4, 9. But, 1 can never be the sum of
two outcomes of a dice. So, we will count the outcomes that make sum of 4 and 9
only.
So, our favorable events will be, (1,3), (2,2), (3,1), (3,6), (4,5),
(5,4), (6,3). These are total 7 events.
P̅ =
P = 1 - `\frac{7}{36}` = `\frac{29}{36}`
Question 35.) If P is the probability of getting at least one head when a coin is tossed N times. Then find the minimum value of N for P≥0.8.
(A) 2
(B) 4
(C) 6
(D) 3
Answer. (D)
Solution.
If a coin is tossed one time then the probability of getting a head P(H) and a tail P(T) is `\frac{1}{2}`.
Here we will use a relation which is associated with probability. It is,
P(X=r) = nCr × (p)r × (q)n-r
Where r is the number of favorable events, p is the probability of
favorable events and q is the probability of non favorable events.
Now for the at least condition we will use one more relation,
P(X≥r+1) = 1 – P(X=r) = 1 – nCr × (p)r × (q)n-r
Now in this question, r=0, n = N, p = `\frac{1}{2}` and q = `\frac{1}{2}`. So,
P(X≥1) = 1 – P(X=0)
= 1 – nC0
= 1 – `\frac{1}{2^{N}}`
Now as per the question,
P(X≥1) ≥ 0.8
1 – (`\frac{1}{2^{N}}`) ≥ 0.8
On solving it we get N ≥ 3.
Question 36.) A six faced die is a biased one. It is twice more likely to show an odd number than to show an even number. It is thrown twice. The probability that the sum of the numbers in two throws is even, is
(A) `\frac{1}{3}`
(B) `\frac{2}{3}`
(C) `\frac{5}{9}`
(D) `\frac{11}{36}`
Answer. (C)
Solution.
Let P is the probability of getting an even number. Then by hypothesis the probability of getting an odd number will be 2P.
Since the events of getting an even number and an odd number are mutually
exclusive and exhaustive. So,
P + 2P = 1
3P = 1
P =
Thus, the probability of getting an odd number in a single throw is =
And the probability of getting an even number in a single throw is =
If the die is thrown twice, then the sum of the
numbers of the outcomes in two throws will be even if both the numbers are either
even or odd.
Hence the required probability = `\frac{2}{3}` × `\frac{2}{3}` + `\frac{1}{3}` × `\frac{1}{3}` = `\frac{5}{9}`
Question 37.) If two fair dices are rolled together and the product of the numbers on the faces is observed then find the probability that their product is an even number.
(A) `\frac{17}{36}`
(B) `\frac{3}{4}`
(C) `\frac{23}{36}`
(D) `\frac{2}{3}`
Answer. (B)
Solution.
When two dices are rolled then the total number of events will be = 6 × 6 = 36
Let P is the probability of getting the product as an even number and P̅
is the probability of getting the product as an odd number. Then,
P = 1 – P̅
Now the product of two numbers will be even if either one of them is
even or both of them are even. Otherwise the product is an odd number. Since
there are 3 odd numbers on the dice; 1, 3, and 5.
The number of combination of these three numbers will be = 3 × 3
P̅ = `\frac{9}{36}` = `\frac{1}{4}`
P = 1 – `\frac{1}{4}` = `\frac{3}{4}`
Question 38.) If a leap year selected at random then what is the probability that it will have five Tuesdays in the month of February?
(A) `\frac{2}{7}`
(B) `\frac{1}{7}`
(C) `\frac{3}{7}`
(D) `\frac{4}{7}`
Answer. (B)
Solution.
As February will have 29 days in a leap year. Hence it will have 4 weeks and 1 day. So, every day of the week will appear only four times except the one day. This one day can be any day of the week.
Hence the required probability = `\frac{1}{7}`
Question 39.) Suppose there are two bags B1 and B2. The bag B1 contains 3 white and 2 red balls, and the bag B2 contains only 1 white ball. Then a fair coin is tossed. If head appears then 1 ball is drawn at random from B1 and placed into B2and if tail appears then 2 balls are drawn at random from B1 and placed into B2. Now, 1 ball is drawn at random from B2. Then the probability of the drawn ball from B2 being white is
(A) `\frac{23}{30}`
(B) `\frac{13}{30}`
(C) `\frac{19}{30}`
(D) `\frac{11}{30}`
Answer. (A)
Solution.
Probability of getting a head when coin is tossed, P(H) = `\frac{1}{2}`
Probability of getting a tail when coin is tossed, P(T) =
Probability of drawing a white ball from bag B1 is, P(W1) =
Probability of drawing a red ball from bag B1 is, P(R1) =
Probability of drawing two white balls from bag B1 is, P(W2) =
Probability of drawing two red balls from bag B1 is, P(R2) =
Probability of drawing a white ball and a red ball from bag B1 is, P(WR)
=
Now we will calculate the probabilities of drawing a ball from the bag
B2. Here, the first case will be if head appears when coin is tossed. Then
after placement of 1 ball from B1 to B2 one ball is drawn from B2.
The probability that it is white if the ball placed from B1 is white is = `\frac{2}{2}` = 1
And the probability that it is white if the ball placed from B1 is red
is =
Hence, the probability of drawing a white ball from B2 if head appears
when coin is tossed is
= `\frac{1}{2}` × `\frac{3}{5}` × 1 + `\frac{1}{2}` × `\frac{2}{5}` × `\frac{1}{2}` = `\frac{2}{5}`
Now, in the second case if tail appears when coin is tossed. Then after
placement of 2 balls from B1 to B2 one ball is drawn from B2.
The probability that it is white if the balls placed from B1 are white is = `\frac{3}{3}` = 1
And the probability that it is white if the balls placed from B1 are red
is =
And the probability that it is white if the balls placed from B1 are 1
white and 1 red is =
Hence, the probability of drawing a white ball from B2 if head appears
when coin is tossed is
= `\frac{1}{2}` × `\frac{3}{10}` × 1 + `\frac{1}{2}` × `\frac{1}{10}` × `\frac{1}{3}` + `\frac{1}{2}` × `\frac{3}{5}` × `\frac{2}{3}` = `\frac{11}{30}`
Hence, the required probability is = `\frac{2}{5}` + `\frac{11}{30}` = `\frac{23}{30}`
Question 40.) Three numbers are chosen at random, one after another with replacement, from the set S = {1, 2, 3, ....., 100}. Then find the probability that all the three numbers are greater than 75.
(A) `\frac{1}{3}`
(B) `\frac{3}{4}`
(C) `\frac{27}{64}`
(D) `\frac{37}{64}`
Answer. (D)
Solution.
Let P is the probability of choosing three numbers greater than 75 and P̅ is the probability of choosing three numbers less than or equal to 75. Then,
P = 1 – P̅
Now, the probability of choosing one number less than or equal to 75 is = `\frac{75}{100}` = `\frac{3}{4}`
P̅ = `\frac{3}{4}` × `\frac{3}{4}` × `\frac{3}{4}` = `\frac{27}{64}`
P = 1 – P̅ = 1 – `\frac{27}{64}` = `\frac{37}{64}`
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