2
Q:

# A bag contains 8 red and 4 green balls. Find the probability that two balls are red and one ball is green when three balls are drawn at random.

 A) 56/99 B) 112/495 C) 78/495 D) None of these

Explanation:

$\inline n(S)=^{12}\textrm{C}_{4}=495$

$\inline n(E)=^{8}\textrm{C}_{2}\times ^{4}\textrm{C}_{1}=112$

$\inline \therefore P(E)=\frac{112}{495}$

Q:

Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 4 or 15 ?

 A) 6/19 B) 3/10 C) 7/10 D) 6/17

Explanation:

Here, S = {1, 2, 3, 4, ...., 19, 20}=> n(s) = 20
Let E = event of getting a multiple of 4 or 15
=multiples od 4 are {4, 8, 12, 16, 20}
And multiples of 15 means multiples of 3 and 5
= {3, 6 , 9, 12, 15, 18, 5, 10, 15, 20}.
= the common multiple is only (15).
=> E = n(E)= 6
Required Probability = P(E) = n(E)/n(S) = 6/20 = 3/10.

4 50
Q:

Out of sixty students, there are 14 who are taking Economics and 29 who are taking Calculus. What is the probability that a randomly chosen student from this group is taking only the Calculus class ?

 A) 8/15 B) 7/15 C) 1/15 D) 4/15

Explanation:

Given total students in the class = 60
Students who are taking Economics = 24 and
Students who are taking Calculus = 32
Students who are taking both subjects = 60-(24 + 32) = 60 - 56 = 4
Students who are taking calculus only = 32 - 4 = 28
probability that a randomly chosen student from this group is taking only the Calculus class = 28/60 = 7/15.

4 43
Q:

Three unbiased coins are tossed. What is the probability of getting at most two heads ?

 A) 4/3 B) 2/3 C) 3/2 D) 3/4

Explanation:

Let S be the sample space.
Here n(S)= $\inline \fn_jvn \small 2^{3}$= 8
Let E be the event of getting atmost two heads. Then,
n(E) = {(H,T,T), (T,H,T), (T,T,H), (H,H,T), (T,H,H), (H,T,H)}
Required probability = n(E)/n(S) = 6/8 = 3/4.

4 122
Q:

Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5 ?

 A) 2/3 B) 1/2 C) 7/8 D) 4/5

Explanation:

Multiples of 3 below 20 are 3, 6, 9, 12, 15, 18
Multiples of 5 below 20 are 5, 10, 15, 20
Required number of possibilities = 10
Total number of possibilities = 20
$\fn_jvn&space;\small&space;\therefore$ Required probability = 10/20 = 1/2.

1 90
Q:

A letter iws takenout at random from 'ASSISTANT'  and another is taken out from 'STATISTICS'. The probability that they are the same letter is :

 A) 35/96 B) 19/90 C) 19/96 D) None of these

Explanation:

$\inline ASSISTANT\rightarrow AA\: I\: N\: SSS\: TT$

$\inline STATISTICS\rightarrow A\: II\: C\: SSS\: TTT$

Here N and C are not common and same letters can be A, I, S, T. Therefore

Probability of choosing A =  $\inline \frac{^{2}\textrm{C}_{1}}{^{9}\textrm{C}_{1}}\times \frac{^{1}\textrm{C}_{1}}{^{10}\textrm{C}_{1}}$ = 1/45

Probability of choosing I = $\inline \frac{1}{^{9}\textrm{C}_{1}}\times \frac{^{2}\textrm{C}_{1}}{^{10}\textrm{C}_{1}}$ = 1/45

Probability of choosing S = $\inline \frac{^{3}\textrm{C}_{1}}{^{9}\textrm{C}_{1}}\times \frac{^{3}\textrm{C}_{1}}{^{10}\textrm{C}_{1}}$ = 1/10

Probability of choosing T = $\inline \frac{^{2}\textrm{C}_{1}}{^{9}\textrm{C}_{1}}\times \frac{^{3}\textrm{C}_{1}}{^{10}\textrm{C}_{1}}$ = 1/15

Hence, Required probability = $\inline \frac{1}{45}+\frac{1}{45}+\frac{1}{10}+\frac{1}{15}=\frac{19}{90}$