14
Q:

# A bag contains 4 red and 3 black balls. A second bag contains 2 red and 4 black balls. One bag is selected at random. From the selected bag, one ball is drawn. Find the probability that the ball drawn is red.

 A) 23/42 B) 19/42 C) 7/32 D) 16/39

Explanation:

A red ball can be drawn in two mutually exclusive ways

(i) Selecting bag I and then drawing a red ball from it.

(ii) Selecting bag II and then drawing a red ball from it.

Let E1, E2 and A denote the events defined as follows:

E1 = selecting bag I,

E2 = selecting bag II

A = drawing a red ball

Since one of the two bags is selected randomly, therefore

P(E1) = 1/2  and  P(E2) = 1/2

Now, $\inline P(\frac{A}{E1})$ = Probability of drawing a red ball when the first bag has been selected = 4/7

$\inline P(\frac{A}{E2})$  = Probability of drawing a red ball when the second bag has been selected = 2/6

Using the law of total probability, we have

P(red ball) = P(A) = $\inline P(E1)P(\frac{A}{E1})+P(E2)P(\frac{A}{E2})$

= $\inline \frac{1}{2}\times \frac{4}{7}+\frac{1}{2}\times \frac{2}{6}=\frac{19}{42}$

Q:

The Manager of a company accepts only one employees leave request for a particular day. If five employees namely Roshan, Mahesh, Sripad, Laxmipriya and Shreyan applied for the leave on the occasion of Diwali. What is the probability that Laxmi priya’s leave request will be approved ?

 A) 1 B) 1/5 C) 5 D) 4/5

Explanation:

Number of applicants = 5
On a day, only 1 leave is approved.
Now favourable events =  1 of 5 applicants is approved
Probability that Laxmi priya's leave is granted = 1/5.

3 22
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 66
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.

6 86
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.

4 140
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

$\fn_jvn&space;\small&space;\therefore$ Required probability = 10/20 = 1/2.