**Question 1:** If a card is picked up random from the a pack of 52 cards. Find the probability that it is a king or queen. **Solution :** In a pack of 52 cards, 4 kings and 4 queens present.

Hence, probability of getting a king or queen = 8/52

= 2/13

**Question 2:** In a throw of 2 dice, find the probability of getting one odd number and one even number. **Solution :** Total number of outcomes in throw of 2 dice = 36

Number of outcomes when one number is odd and one is even {(1, 2)(1, 4)(1, 6), (2, 1)(2, 3)(2, 5), (3, 2)(3, 4)(3, 6), (4, 1)(4, 3)(4, 5), (5, 2)(5, 4)(5, 6), (6, 1)(6, 3)(6, 5)}

Hence, Required probability = 18/36 = 1/2

**Question 3:** What will be the probability that a leap year chosen at random will have 53 Sundays. **Solution :** A leap year has 366 days.

To find number of weeks = 366/7 = 52 weeks complete

2 days left in the year either of them can be Sunday.

(Saturday or Sunday), (Sunday or Monday)

So possibility of 53 Sundays in a leap year = 2/7

**Question 4:** There are 5 women and 3 men applicants for a job.Only two out of eight are selected for a job.The probability that at least one of the selected person will be a women is: **Solution :** Selection can be done like that

First is a woman and second is a man

OR first is a man and second is a woman

OR both woman

Required probability = (5/8)(3/7) + (3/8)(5/7) + (5/8)(4/7)

= 15/56 + 15/56 + 20/56

= 50/56

= 25/28* ***Alternate Explanation**** 1**

Find the probability of men i-e

3C2 / 8C2 = 3/28

Now , the probability of women selected will be

1 – p(men) = 1 – 3/28 = 25/28

**Alternate Explanation 2**

Number of favourable outcomes = 5C2 + (5C1 x 3C1)= 25

Total number of outcomes = 8C2 = 28

Therefore, required probability = 25 / 28

**Question 5:** The probability that A can solve the problem is 3/4 and B can solve the problem is 4/5. If both of them attempt the problem, then what is the probability that problem gets solved. **Solution :** Probability that A cannot solve the problem = 1/4

and Probability that B cannot solve the problem = 1/5

Probability that problem not solved = 1/4 x 1/5 = 1/20

Hence, the problem is solved either by A or B = 1 – 1/20

= 19/20

**Question 6:** 200 students appeared for GATE and CAT examinations. 60% passed in GATE, 40% passed the CAT and 25% passed both. Find the probability that a student selected at random has failed in both the examinations? **Solution :** Number of students passed in GATE = 200 x 60% = 120

Number of students passed in CAT = 200 x 40% = 80

Number of students passed in both = 200 x 25% = 50

Number of students passed in either GATE or CAT = 120 + 80 – 50 = 150

Hence, Number of students failed in both = 200 – 150 = 50

Required probability = 50/200 = 1/4

**Question 7:** A box contains 40 bulbs out of which 4 are defective.Two bulbs are selected at random from the box. What will be the probability that both bulbs found to be defective? **Solution :** Both bulbs should come from the defective bulbs without replacement.

Required probability = 4/40 x 3/39 = 1/130

**Question 8:** Ten persons are seated round a circular table. What is the probability that three friends always sit together? **Solution :** Total number of ways = 9!

Total number of ways in which two people sit together = 7! x 3!

Required probability = 7! x 3!/ 9!

**Question 9:** A bag contains pens numbered from 1 to 17. A pen is drawn and replaced. Then one more pen is drawn and replaced. What will be the probability that first pen drawn is even and second one is odd. **Solution :** In first draw, we have 8 even numbered pens out of 15 and in second we have 9 odd numbered pens.

Required probability = 8/17 x 9/17

= 72/289

**Question 10:** If P(A)=2/3, P(B)=1/4, P(A ∩ B)=1/3 then find the P(A ∪ B) **Solution:** P(A ∪ B)= P(A) + P(B) – P(A ∩ B)

=> 2/3 + 1/4 – 1/3

=> (8 + 3 – 4)/12

=> 7/12