## Wipro Practice Probability Questions

Dear Reader,

Below are three probability questions which require you to calculate probabilities for consecutive events based on given conditions.

**Question 1**

In an exhibition arranged every year in island Grounds, there is a toy train with twelve different whistles . It was displayed as if travelling in a circle of radius of 8 m at 12 m per minute. But on account of a defect in the toy it was making only two whistles at random. What are the odds that this toy train makes three consecutive music whistle of the same type?

a)1/8 b)1/16 c)1/32 d)none of these.

**Answer :** a) 1/8

Solution :

Originally there are twelve different whistles. But on account of defect only two whistles are now being made. Therefore probability of one particular whistle to blow would be 1/2.

Probability of same whistle (out of the two available now) being made three times consecutively is = (Probability of one particular whistle blowing)^{3 } = 1/2 X 1/2 X 1/2 = 1/8.

**Question 2**

Costco Toy Manufacturers made a new toy which is capable of making 8 different sounds as the toy is made to run with a battery being fitted to it. The toy is capable of being made to run in a circular way and also in straight line. If the toy goes and hits another surface it will turn and run again. This toy is capable of running continuously even for 30 minutes when fitted with high value batteries. On account of a defect now the battery was making only three different sounds. What are the odds that this toy makes two consecutive sounds of one type and final sound of any other type ?

a) 1/9 b) 1/27 c) 2/27 d) none of these.

**Answer : **b) 2/27

Solution :

The toy has 8 different sounds fitted. But on account of defect it can make only 3 different sounds now.

P1 = Probability that the toy makes two consecutive sounds of same type = (Probability of the toy making any one particular sound)^{2}

Probability for the same toy to make the third sound that is different from first = P1 X Probability of the third sound being a different one -> eq 1

Now, after a toy makes a particular sound two times, the probability of the third sound being a different one would be : No of other possible sounds / Total number of sounds

Since there are totally 3 sounds and one sound has been played consecutively for the first two times, the number of other possible sounds is 3 - 1 = 2. Applying this in above formula we get

Probability of the third sound being a different one = No of other possible sounds / Total number of sounds = 2/3.

Applying the above value in eq 1 we get

Probability for the same toy to make the third sound that is different from first = P1 X 2/3 = (1/3)^{2}X2/3 = 2/27

** Question 3**

Emergent Cellphone company made a cell phone which is capable of giving different caller tunes. It had originally 10 different caller tunes. This cell phone fell into a box containing water and it resulted in damage and now the cell phone is making only any one of 2 different caller tunes when calls are received. What is the probability that the cell phone will ring both the caller tunes at the same instance ?

a) 1/2 b) 1 c) 1/3 d) none of these.

**Answer : ** d) none of these

Solution :

This is a tricky question. In the question it is clearly mentioned that at any instance any one of the two caller tunes will be played. There is absolutely no hint that two caller tunes can be played at the same instance. Hence two caller tunes being played at the same instance is an impossible case and hence the probability should be 0.

Hence, none of these will be the answer.

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