Probability 2

Q1. In Formula 1 race, 18 cars were numbered 1 to 18. Probability that car 1 will win is 1/6 and the car 2 will win is 1/10 and that car 3 will win is 1/8.
Assuming that tie is impossible then find the chance that one of the three will win?
a. 119/120
b. 47/120
c. 11/129
d. 1/5

Q2. A dice is thrown (2a + 1) times, where ‘a’ is a natural number. The probability that faces with even numbers shown off number of times is :
a. (2n + 1)/(4n + 3)
b. Less than 1/2
c. Greater than 1/2
d. None of these

Q3. A shooter firing at a target has 10% chance of hitting the target in one shot. The number of times he must fire at the target to have above 50% chance of hitting the target is
a. 11
b. 9
c. 7
d. 5

Q4. On a chess board, three squares are chosen at random. What is the probability that they are in a diagonal line?
a. 7/1488
b. 3/3472
c. 11/ 32
d. 7/744

Q5. Vikas and Sharad appear for an interview for two vacant posts in an office. The probability of Vikas selection is 3/7 and that of Sharad’s selection is 2/3. what is the probability that only one of them is selected?
a. 11/21
b. 2/7
c. 8/21
d. 1/7

Q6. 100 students were chosen for survey. It was found 60 like their school days, 50 like their college days and 30 like both. Find the probability that a student selected at random doesn’t liked either of his days?
a. 1/3
b. 1/5
c. 1/6
d. None of these
 

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Q6. 100 students were chosen

Q6. 100 students were chosen for survey. It was found 60 like their school days, 50 like their college days and 30 like both. Find the probability that a student selected at random doesn’t liked either of his days?
a. 1/3
b. 1/5
c. 1/6
d. None of these

student doen't like any days=20

so probability=20/100
=1/5

ans is (b)

Little Star

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Q5. Vikas and Sharad appear

Q5. Vikas and Sharad appear for an interview for two vacant posts in an office. The probability of Vikas selection is 3/7 and that of Sharad’s selection is 2/3. what is the probability that only one of them is selected?
a. 11/21
b. 2/7
c. 8/21
d. 1/7

since both are mutually dependent events.
sp prob of any one of them selection is wen only one get selected is.
prob of 1 got selected bt 2 is not + prob of 2 got selected bt 1 is not
(add them as they are mutually exclusive)
so prob is 3/7 * 1/3 + 4/7 * 2/3 is 11 /21.

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Q4. On a chess board, three

Q4. On a chess board, three squares are chosen at random. What is the probability that they are in a diagonal line?

for solution of the above qs i would like to draw a chess board to make it clear about diagonals.
& it miror immage to clearify other set of diagonal.

1 2 3 4 5 6 7 8          8 7 6 5 4 3 2 1
2 3 4 5 6 7 8 7          7 8 7 6 5 4 3 2
3 4 5 6 7 8 7 6          6 7 8 7 6 5 4 3
4 5 6 7 8 7 6 5          5 6 7 8 7 6 5 4
5 6 7 8 7 6 5 4          4 5 6 7 8 7 6 5
6 7 8 7 6 5 4 3          3 4 5 6 7 8 7 6
7 8 7 6 5 4 3 2          2 3 4 5 6 7 8 7
8 7 6 5 4 3 2 1          1 2 3 4 5 6 7 8 

So diagonals are having same number bt we have to take 3 at a time so position for 1 & 2 diagonal is ruled out.
now the required probability is
2* ( 2( 3C3 + 4C3 + 5C3 + 6C3 + 7C3) + 8C3) / 64C3

i.e. 7/744.

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Q3. A shooter firing at a

Q3. A shooter firing at a target has 10% chance of hitting the target in one shot. The number of times he must fire at the target to have above 50% chance of hitting the target is
a. 11
b. 9
c. 7
d. 5

Required prob is
1/10+ 9/10*1/10 + (9/10)2 *1/10 + ................ + (9/10)n *1/10 > 1/2

gives geometric series so we have on solving with GP help.

(9/10)n < 1/2
gives n= 7

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Q1. In Formula 1 race, 18

Q1. In Formula 1 race, 18 cars were numbered 1 to 18. Probability that car 1 will win is 1/6 and the car 2 will win is 1/10 and that car 3 will win is 1/8.
Assuming that tie is impossible then find the chance that one of the three will win?
a. 119/120
b. 47/120
c. 11/129
d. 1/5

Req prob is 1/8+1/6+1/10 i.e. 47/120

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Need Comments

Q2. A dice is thrown (2a + 1) times, where ‘a’ is a natural number. The probability that faces with even numbers shown off number of times is :
a. (2n + 1)/(4n + 3)
b. Less than 1/2
c. Greater than 1/2
d. None of these

throwing one dice (2n+1) times or throwing (2n+1) dices at same time gives same probability of an occurence of an event.
The probability that faces with even numbers with one dice is 1/2.
so for (2n+1) dices it is

(1/2)(2n+1) < 1/2 for any natural number n.

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