1
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 that the card  drawn is neither a spade nor a king.

 A) 0 B) 9/13 C) 1/2 D) 4/13

Explanation:

There are 13 spades ( including one king). Besides there are 3 more kings in remaining 3 suits

Thus   $\inline n(E)=13+3=16$

Hence $\inline n(\overline{E})=52 -16=36$

Therefore, $\inline P(\overline{E})=\frac{36}{52}=\frac{9}{13}$

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 = $\inline \frac{1}{45}+\frac{1}{45}+\frac{1}{10}+\frac{1}{15}=\frac{19}{90}$

44 994
Q:

8 couples (husband and wife) attend a dance show "Nach Baliye' in a popular TV channel ; A lucky draw in which 4 persons picked up for a prize is held, then the probability that there is atleast one couple will be selected is :

 A) 8/39 B) 15/39 C) 12/13 D) None of these

Explanation:

P( selecting atleast one couple) = 1 - P(selecting none of the couples for the prize)

= $\inline 1-\left ( \frac{^{16}\textrm{C}_{1}\times^{14}\textrm{C}_{1}\times ^{12}\textrm{C}_{1}\times ^{10}\textrm{C}_{1} }{^{16}\textrm{C}_{4}} \right )=\frac{15}{39}$

36 1520
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) = $\inline P(E1)P(\frac{A}{E1})+P(E2)P(\frac{A}{E2})$

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

9 1105
Q:

Ajay and his wife Reshmi appear in an interview for two vaccancies in the same post. The Probability of Ajay's selection is 1/7 and that of his wife Reshmi's selection is 1/5. What is the probability that only one of them will be selected?

 A) 5/7 B) 1/5 C) 2/7 D) 2/35

Explanation:

P( only one of them will be selected)

= p[(E and not F) or (F and not E)]

= $\inline P[(E\cap \bar{F})\cup (F\cap \bar{E})]$

= $\inline P(E)P(\bar{F})+P(F)P(\bar{E})$

=  $\inline \frac{1}{7}\times \frac{4}{5}+\frac{1}{5}\times \frac{6}{7}=\frac{2}{7}$

11 723
Q:

The probabilities that a student will receive an A, B, C or D grade are 0.4, 0.3 , 0.2 and 0.1 respectively. Find the probability that a student will receive Atleast B grade.

 A) 0.21 B) 0.3 C) 0.7 D) None of these

Explanation:

P(atleast B) = P( B or A) = P(B) + P(A) = (0.3) + (0.4) = 0.7