**Two cubes have their face painted as either red or blue color. 1st cube has 5 faces red and 1 face blue. The Probability that both the faces show same color on rolling is 12. How many red faces are present on the 2nd cube.**

__Question:__**Options:**

A. | 6 | |

B. | 4 | |

C. | 3 | |

D. | 2 |

**Answer: Option C**

: C

Suppose x faces of cube 2 is red then 6-x will be blue. P(RR or BB)= 56×x6+16×(6−x)6=12 5x36+(6−x)36=12 4x+6=18 x=3 Hence number of red faces is 3.

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## More Questions on This Topic :

**The probability of happening an event A is 0.5 and that of B is 0.3. If A and B are mutually exclusive events, then the probability of happening neither A nor B is**

__Question 1.__- 0.6
- 0.2
- 0.21
- None of these

**Answer: Option B**

: B

P(¯¯¯¯A∩¯¯¯¯B)=P(¯¯¯¯A∪¯¯¯¯B)−1−P(A∪B)

Since A and B are mutually exclusive,

P(A∪B)=P(A)+P(B)

Hence required probability =1–(0.5+0.3)=0.2

**Of the 10 prizes 5 prizes are of category Platinum 3 of gold and 2 of silver and they are placed in an enclosure for an olympiad contest. The prizes are awarded by allowing winners to select randomly from the prizes remaining. When the 8th participant goes to collect the prize what the probability that last 3 prizes are 1 of platinum 1 of gold and 1 of silver?**

__Question 2.__- 15
- 14
- 13
- 110

**Answer: Option B**

: B

Sample space: 10C7. because 7 prizes have already been selected= 120. Favorable: 5C4×3C2×2C1=30. P(E)= 30120=14

**Urn A contains 6 red and 4 black balls and urn B contains 4 red and 6 black balls. One ball is drawn at random from urn A and placed in urn B. Then one ball is drawn at random from urn B and placed in urn A. If one ball is now drawn at random from urn A, the probability that it is found to be red, is**

__Question 3.__- 3255
- 2155
- 1955
- None of these

**Answer: Option A**

: A

Let the events are

R1 = A red ball is drawn from urn A and placed in B

B1 = A black ball is drawn from urn A and placed in B

R2 = A red ball is drawn from urn B and placed in A

B2 = A black ball is drawn from urn B and placed in A

R = A red ball is drawn in the second attempt from A

Then the required probability

= P(R1R2R)+(R1B2R)+P(B1R2R)+P(B1B2R)

= P(R1)P(R2)P(R)+P(R1)P(B2)P(R)+P(B1)P(R2)P(R)+P(B1)P(B2)P(R)

= 610×511×610+610×611×510+410×411×710+410×711×610

= 3255

**If the integers m and n are chosen at random between 1 and 100, then the probability that a number of the form 7m+7n is divisible by 5 equals**

__Question 4.__- 14
- 17
- 18
- 149

**Answer: Option A**

: A

Since m and n are selected between 1 and 100, hence sample space = 100×100.

Also 71=7,72=49,73=343,74=2401,75=16807 etc. Hence 1, 3, 7 and 9 will be the last digits in the powers of 7. Hence for favorable cases

nm→↓1,11,21,3....1,1002,12,22,3...2,100...........................................100,1100,2100,3100,100

For m = 1; n = 3, 7, 11….. 97

∴Favorable cases = 25

Similarly for every m, favorable n are 25.

∴Total favourable cases = 100×25

Hence required probability = 100×25100×100=14

**A fortune teller has numbers from 1-7 in his box. He draws 3 numbers from it. What is the probability that the number picked is of alternate order as odd-even-odd or even-odd-even?**

__Question 5.__- 114
- 27
- 914
- None of the above.

**Answer: Option B**

: B

Sample Space: 7×6×5=210 odd-even-odd= 4×3×3=36 even-odd-even= 3×4×2=24 Total ways: 60 P(E)= 60210=27

**The probability that a 'P and C' question will be asked in IIT JEE is 25 and probability question is uploaded is 47. If the probability of getting at least 1 is 23 what is the probability that questions from both the topics are asked.**

__Question 6.__- 17105
- 16105
- 135
- 635

**Answer: Option A**

: A

P(A U B)= P(A)+ P(B)- P(A intersection B) 23=(25)+(47)−x x=17105

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