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**Practice Aptitude Questions (Probability) Day-152**

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**1) Two friends Harish and Kalyan appeared for an exam. Let A be the event that Harish is selected and B is the event that Kalyan is selected. The probability of A is 2/5 and that of B is 3/7. Find the probability that both of them are selected.**

a) 35/36

b) 5/35

c) 5/12

d) 6/35

e) None of these

**2) Preethi was born between August 26 ^{th} and 30^{th} (26^{th} and 30^{th} excluding). Her year of birth is also unknown. What is the probability of Preethi being born on a Monday?**

a) 4/7

b) 1

c) 0

d) 3/7

e) None of these

**3) A number is selected at random from first 40 natural numbers. What is the chance that it is a multiple of either 4 or 14?**

a) 1/5

b) 11/40

c) 3/10

d) 10/40

e) None of these

**4) Srinaya forgot the last digit of an 11 digit land line number. If she randomly dials the final 2 digits after correctly dialing the first nine, then what is the chance of dialing the correct number?**

a) 1/10

b) 1/100

c) 1/1000

d) 1/11

e) None of these

**5) A military man can strike a target once in 3 bullets. If he fires 3 bullets in succession, what is the probability that he will strike his target?**

a) 1/3

b) 19/27

c) 2/3

d) 8/27

e) None of these

**6) Daniel speaks truth in 2/5 cases and Sherin lies in 3/7cases. What is the percentage of cases in which both Daniel and Sherin contradict each other in stating a fact?**

a) 6%

b) 4%

c) 3%

d) 5%

e) None of these

**7) The probability that a bullet fired from a point will strike the target is 3/4. 5 such bullets are fired simultaneously towards the target from that very point. What is the probability that the target will be strike?**

a) 255/256

b) 1/256

c) 251/256

d) 1/64

e) None of these

**8) Akshaya and Nikitha play a game where each is asked to select a number from 1 to 8. If the two numbers same, both of them win a prize. The probability that they will not win a prize in a single trial is:**

a) 1/8

b) 5/8

c) 4/8

d) 7/8

e) None of these

**9) There are three events X,Y and Z, one of which must and only can happen. If the odds are 7:4 against X, 5:3 against Y, the odds against Z must be:**

a) 65/23

b) 47/51

c) 27/65

d) 37/53

e) None of these

**10) Six male and five female stands in queue for buy a cinema ticket. The probability that they stand in alternate positions is:**

a) 1/258

b) 1/462

c) 1/365

d) 1/512

e) None of these

**Answers:**

**1) Answer: D**

Given, A be the event that Harish is selected and

B is the event that Kalyan is selected.

P(A)= 2/5

P(B)=3/7

Let C be the event that both are selected.

P(C)=P(A)×P(B) as A and B are independent events:

P(C) = 2/5*3/7

P(C) =6/35

The probability that both of them are selected is 6/35

**2) Answer: D**

Since the year of birth is Unknown, the birthday being on Monday can have a zero probability.

Also since between 26th and 30th, there are three days

i.e. 27th, 28th and 29^{th}

We have following possibilities on these three dates,

Monday, Tuesday, Wednesday

Tuesday, Wednesday, Thursday

Wednesday, Thursday, Friday

Thursday, Friday, Saturday,

Friday, Saturday, Sunday

Saturday, Sunday, Monday

Sunday, Monday, Tuesday

Out of these 7 events, we have 3 chances of his birthday falling on Monday

Probability = favourable events/ total events

Probability = 3/7

Therefore, the probability of birthday falling on Monday can be 3/7.

**3) Answer: B**

We know that,

Probability=Favorable Cases/Total Cases

The probability that the number is a multiple of 4 is 10/40

Since favorable cases here {4,8,12,16,20,24,28,32,36,40}

=10 cases

Total cases= 40 cases

Similarly the probability that the number is a multiple of 14 is 2/40.

Since favorable cases here {14,28}

=2 cases

Total cases= 40cases

Multiple of 4 or 14 have common multiple from 1 to 40 is 28.

Hence, these events are mutually exclusive events.

Therefore chance that the selected number is a multiple of4 or 14 is:

=(10+2-1)/40

=11/40

**4) Answer: B**

It is given that last two digits are randomly dialed.

Then each of the digits can be selected out of 10 digits in 10 ways.

Hence required probability

=(1/10)^{2}

=1/100

**5) Answer: B**

The military man will strike the target even if he strike it once or twice or all three times in the three bullets that he takes.

So, the only case where the man will not strike the target is when he fails to strike the target even in one of the three bullets that he takes.

The probability that he will not strike the target in one bullets =1 – Probability that he will strike target in exact one bullets

=1-1/3

=2/3

Therefore, the probability that he will not strike the target in all the three bullets

=2/3*2/3*2/3

=8/27

Hence, the probability that he will strike the target at least in one of the four bullets:

=(1-8/27)

=19/27

**6) Answer: B**

Daniel and Sherin will contradict each other when one speaks truth and other speaks lies.

Probability of Daniel speak truth and Sherin lies

=2/5*3/7

=6/35

Probability of Sherin speak truth and Daniel lies

=4/7*3/5

=12/35

The two probabilities are mutually exclusive.

Hence, probabilities that Daniel and Sherin contradict each other:

=6/35 +12/35

=18/35

=18/35*100

=51.4%

**7) Answer: A**

Probability of a bullet not striking the target = ¼

Probability that none of the 5 bullets will strike the target

=(1/4)^{5}

=1/256

Probability that the target will strike at least once:

=1-1/256

=255/256

**8) Answer: D**

Total number of ways in which both of them can select a number each:

=8*8

=64

Total number of ways in which both of them can select a same number so that they both can win:

=8ways

[They both can select {(1,1),(2,2),(3,3),(4,4),(5,5),(6,6),(7,7),(8,8)}]Probability that they win the prize:

=Favourable Cases / Total Cases

=8/64

=1/8

Probability that they do not win a prize:

=1-1/8

=7/8

**9) Answer: A**

According to the question,

P(X’)/P(X)=7/4

P(X’)=7/11

P(X)=4/11

P(Y’)/P(Y)=5/3

P(Y’)=5/8,P(Y)=3/8

Now, out of X,Y and Z, one and only one can happen.

P(X)+P(Y)+P(Z)=1

4/11+3/8+P(Z)=1

P(Z)=1-4/11-3/8

=88-32-33/88

=23/88

P(Z’)=1-P(Z)

=1-23/88

=65/88

So odd against z

P(z’)/p(z)=65/23

**10) Answer: B**

Total number of possible arrangements for Six male and five female stand in queue =11!

When they occupy alternate position the arrangement would be like:

MFMFMFMFMFM

Thus, total number of possible arrangements for males,

= 6*5*4*3*2

Total number of possible arrangements for females,

=5*4*3*2

Required probability

=6*5*4*3*2*5*4*3*2/11*10*9*8*7*6*5*4*3*2

= 1/462

**Daily Practice Test Schedule | Good Luck**

Topic | Daily Publishing Time |

Daily News Papers & Editorials | 8.00 AM |

Current Affairs Quiz | 9.00 AM |

Quantitative Aptitude “20-20” | 11.00 AM |

Vocabulary (Based on The Hindu) | 12.00 PM |

General Awareness “20-20” | 1.00 PM |

English Language “20-20” | 2.00 PM |

Reasoning Puzzles & Seating | 4.00 PM |

Daily Current Affairs Updates | 5.00 PM |

Data Interpretation / Application Sums (Topic Wise) | 6.00 PM |

Reasoning Ability “20-20” | 7.00 PM |

English Language (New Pattern Questions) | 8.00 PM |

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