Quiz 3

Question 1

When 4 dice are rolled simultaneously, what is the probability that the same number appears on all of them?

1/234

1/216

1/456

1/652

  • PrepInsta Explanation
  • User Explanation

Solution:
Sample space (Denominator): When 4 dice are thrown simultaneously, then the total number of possible outcomes is 64 = 1296
Event (Numerator): The chances that all the dice show same number (1,1,1,1), (2,2,2,2), (3,3,3,3), (4,4,4,4), (5,5,5,5), (6,6,6,6)} is 6.
Probability = Event/ Sample space = 6/64 = 1/63 = 1/216.

 

Question 2

Ramesh plans to order a birthday gift for his friend from an online retailer. However, the birthday coincides with the festival season during which there is a huge demand for buying online goods and hence deliveries are often delayed. He estimates that the probability of receiving the gift, in time, from the retailers A, B, C and D would be 0.6, 0.8, 0.9 and 0.5 respectively. Playing safe, he orders from all four retailers simultaneously. What would be the probability that his friend would receive the gift in time?

0.997

0.996

0.995

None of these

  • PrepInsta Explanation
  • User Explanation

His friend will receive the gift if at least one of the retailers delivers the gift.
Conversely, his friend will not receive the gift only if all four retailers fail to deliver the gift.
So, probability of receiving the gift = 1 – {probability of none of the retailers delivering the gift}
Probability that none of the retailers deliver
The probability of not receiving the gift from retailer A, P(A’) = 1 - P(A) = 1 - 0.6 = 0.4.
The probability of not receiving the gift from retailer B, P(B’) = 1 – 0.8 = 0.2
The probability of not receiving the gift from retailer C, P(C’) = 1 – 0.9 = 0.1
The probability of not receiving the gift from retailer D, P(D’) = 1 – 0.5 = 0.5


The probability that none of the retailers deliver the gift
= P(A’) * P(B’) * P(C’) * P(D’)
= (0.4 × 0.2 × 0.1 × 0.5)
=0.004
Therefore, the probability of Ramesh’s friend receiving a gift
= {1 - 0.004} = 0.996

Question 3

A man can hit a target once in 4 shots. If he fires 4 shots in succession, what is the probability that he will hit his target?

1/997

124/45

175/256

133/256

  • PrepInsta Explanation
  • User Explanation

Explanatory Answer
The man will hit the target even if he hits it once or twice or thrice or all four times in the four shots that he takes.

So, the only case where the man will not hit the target is when he fails to hit the target even in one of the four shots that he takes.

The probability that he will not hit the target in one shot = 1 - =
Therefore, the probability that he will not hit the target in all the four shots =

Hence, the probability that he will hit the target at least in one of the four shots =
1 - 81/256
= 175/256

Question 4

A number is selected at random from first thirty natural numbers. What is the chance that it is a multiple of either 3 or 13?

2/3

2/5

1/4

1/5

  • PrepInsta Explanation
  • User Explanation

Solution :
The probability that the number is a multiple of 3 is 10/30. (Since 3*10 = 30).

Similarly the probability that the number is a multiple of 13 is 2/30. {Since 13*2 = 26).

Neither 3 nor 13 has common multiple from 1 to 30. Hence these events are mutually exclusive events.
Therefore chance that the selected number is a multiple of 3 or 13 is (10+2)/30 = 2/5

Question 5

An anti aircraft gun can fire four shots at a time. If the probabilities of the first, second, third and the last shot hitting the enemy aircraft are 0.7, 0.6, 0.5 and 0.4, what is the probability that four shots aimed at an enemy aircraft will bring the aircraft down?

0.988

0.964

0.934

None of these

  • PrepInsta Explanation
  • User Explanation

Solution:
The enemy aircraft will be brought down even if one of the four shots hits the aircraft.
The opposite of this situation is that none of the four shots hit the aircraft.
The probability that none of the four shots hit the aircraft is given by (1-0.7)(1-0.6)(1-0.5)(1-0.4) = 0.3*0.4*0.5*0.6 = 0.036

So, the probability that at least one of the four hits the aircraft = 1 – 0.036 = 0.964.

Question 6

An experiment succeeds twice as often as it fails. What is the probability that in the next 5 trials there will be four successes?

[2/3]^4

5∗[2/3]^4∗[1/3]

[2/3]^4∗[1/3]

  • PrepInsta Explanation
  • User Explanation

Solution:
An experiment succeeds twice as often as it fails.
i.e. the probability of its success is 2/3 and the probability of its failure is 1/3.
In the next 5 trials the experiment needs to succeed in 4 out of the 5 trials.
4 out of the 5 trials in which it succeeds could be selected in 5C4 ways = 5 ways.
And as 4 of them are successes, they have a probability of 2/3 and the one that is a failure will have a probability of 1/3.
Hence, the required probability = 5*((2/3)^4)*(1/3)

["0","40","60","80","100"]
["Need more practice!","Keep trying!","Not bad!","Good work!","Perfect!"]

Analytics below shows your performance in various Mocks on PrepInsta

Your average Analytics for this Quiz

Rank

-

Percentile

0%

Start Preparing
Detailed performance

Completed

0/6

Accuracy

0%

Comments