17
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:

A room contains 3 brown, 5 black and 4 white chairs. Two chairs are picked and are put in the lawn. What is the probability that none of the chairs picked is white ?

 A) 14/33 B) 14/55 C) 12/55 D) 13/33

Explanation:

Total number of chairs = (3 + 5 + 4) = 12.

Let S be the sample space.

Then, n(s)= Number of ways of picking 2 chairs out of 12

12×11/2×66

Let n(E) = number of events of selecting 2 chairs for selecting no white chairs.

=> 8C8×7/2×28

Therefore required probability = 28/66 = 14/33.

6 405
Q:

The probability that a bowler bowled a ball from a point will hit by the batsman is ¼. Three such balls are bowled simultaneously towards the batsman from that very point. What is the probability that the batsman will hit the ball ?

 A) 37/64 B) 27/56 C) 11/13 D) 9/8

Explanation:

Probability of not hitting the ball = 1- 1/4 =
Then, the probability that the batsman will hit the ball =

8 398
Q:

One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn find the probability the card drawn is red.

 A) 2/3 B) 1/2 C) 1/52 D) 13/51

Explanation:

We know that:

When one card is drawn from a pack of 52 cards

The numbers of possible outcomes n(s) = 52

We know that there are 26 red cards in the pack of 52 cards
⇒ The numbers of favorable outcomes n(E) = 26

Probability of occurrence of an event: P(E)=Number of favorable outcomes/Numeber of possible outcomes=n(E)/n(S)

∴ required probability = 26/52 = 1/2.

7 498
Q:

A fair coin is tossed 11 times. What is the probability that only the first two tosses will yield heads ?

 A) (1/2)^11 B) (9)(1/2) C) (11C2)(1/2)^9 D) (1/2)

Explanation:

Probability of occurrence of an event,

P(E) = Number of favorable outcomes/Numeber of possible outcomes = n(E)/n(S)

⇒ Probability of getting head in one coin = ½,

⇒ Probability of not getting head in one coin = 1- ½ = ½,

Hence,

All the 11 tosses are independent of each other.

∴ Required probability of getting only 2 times heads =$\inline \fn_jvn \left ( \frac{1}{2} \right )^{2}x\left ( \frac{1}{2}\right ) ^{9}=\left ( \frac{1}{2} \right )^{11}$

5 417
Q:

The first 8 alphabets are written down at random. what is the probability that the letters b,c,d,e always come together ?

 A) 1/7 B) 8! C) 7! D) 1/14

Explanation:

The 8 letters can be written in 8! ways.
n(S) = 8!
Let E be the event that the letters b,c,d,e always come together when the first 8 alphabets are written down.
Now the letters a, (bcde), f, g and h can be arranged in 5! ways.
The letters b,c,d and e can be arranged themselves in 4! ways.
n(E) = 5! x 4!
Now, the required P(E) = n(E)/n(S) = 5! x 4!/8! = 1/14