Probability without Replacement (Dependent Events)

Q.

A lot contains $10$ items out of which $3$ are defective. Three items are chosen from the lot one after another at random without replacement. Find the probability that all three are defective.

Q.

A box contains $20$ electric bulbs, out of which $4$ are defective. One bulb is chosen at random from this box. The probability of getting a defective bulb is :

1. $\frac{4}{5}$

2. $\frac{1}{5}$

3. $\frac{7}{20}$

4. $\frac{11}{20}$

Q. A purse contains four $\$$5 bills, five $\$$10 bills and three $\$$20 bills. Two bills are selected without the first selection being replaced, find probability of getting both $\$5$ bills.

1. $\frac{1}{3}$
2. $\frac{3}{11}$
3. $\frac{1}{11}$
4. $\frac{1}{12}$

Q. The probability of students playing basketball and football and both the games is given below:
What is the probability of students playing a game when they have already played football?

1. $\frac{255}{247}$
2. $\frac{15}{14}$
3. $\frac{247}{255}$
4. $\frac{250}{247}$

Q. In a class test, 8 out of 30 students got full marks. If three students are chosen at random, what is the probability that all three got full marks?

1. $\text{P(E)}=\frac{4}{15}\times \frac{7}{29}\times \frac{3}{14}$
   
2. $\text{P(E)}={\left(\frac{4}{15}\right)}^3$
   
3. $\text{P(E)}=\frac{8}{30}\times \frac{7}{29}\times \frac{7}{28}$
   
4. (a) and (b) above.

Q. A college survey found that 6 out of 24 students walk to college. If 3 students are selected at random without replacement, what is the probability that all 3 walk to college?

1. $\text{P(E)}=\frac{23}{506}$
2. $\text{P(E)}=\frac{5}{506}$
3. $\text{P(E)}=\frac{1}{506}$
4. $\text{P(E)}=\frac{1}{605}$

Q. 16 people are their in a college cricket team. 6 of them are batsman and 10 of them are bowlers. Two people are chosen at random. Calculate the probability that first is a batsman and second is bowler(without replacement).

1. $\text{P(A and B)}=\frac{1}{4}$
2. $\text{P(A and B)}=\frac{3}{8}$
3. $\text{P(A and B)}=\frac{2}{3}$
4. $\text{P(A and B)}=\frac{3}{4}$

Q. A bag contains 20 green marbles and 12 blue marbles. Four marbles are drawn at random. What is the probability of getting all green marbles?

1. $\text{P(E)}=\frac{95}{248}\times \frac{51}{145}$
2. $\text{P(E)}=\frac{323}{248}\times \frac{51}{145}$
3. $\text{P(E)}=\frac{15}{248}\times \frac{95}{145}$