(2)     3/13

(4)

Probability of loosing the game

= 1 – Probability of winning the game

= 1 – 0.79 = 0.21

(2)

Given data : 1, 4, 7, 9, 16, 21, 25

After removing even numbers data is: 1, 7, 9, 21, 25

Prime number : 7

Thus, required probability = 1/5

(3)

Total possible number of outcomes = 36

Number of outcomes in which 6 will come up at least once = 11

$\therefore$ Required probability $=\frac{11}{36}$

(1)

When two dice are tossed together,

Total possible outcomes = 62 = 36

No. of cases getting sum two : {(1, 1)}

No. of cases getting sum three : {(1, 2)(2, 1)}

No. of cases getting sum five : {(1, 4)(4, 1), (2, 3), (3, 2)}

Thus total cases getting sum 2, 3 or 5 = 7

Therefore, required probability = 7/36

(1)       1/10

(1)        33

(1)      7/36

(4)

Total number of outcomes = 36

Sum of two numbers to be more than 10 = {(6, 5) (5, 6) (6, 6)}

Required Probability = 3/36 = 1/12

(3)

Given, A box contains card numbered 6 to 55.

Total number of cards = 50

Perfect square number are = 9, 16, 25, 36, 49

$\therefore$ Required Probability = 5/50 = 1/10

(3)

Given numbers are 1, 2, 3, …, 15.

∴ The number of possible outcomes = 15

The outcomes favourable to event E ‘getting a multiple of 4’ are 4, 8, 12.

∴ Number of favourable outcomes of event E = 3

∴ Probability that a number selected randomly is a multiple of $4 = 3/15$

(2)

When two dice are thrown together, the possible outcomes are:

$(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),$

$(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)$

$(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),$

$(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),$

$(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),$

$(6,1),(6,2),(6,3),(6,4),(6,5),(6,6).$

Therefore, the total number of possible outcomes = 36 $\Rightarrow n\left(S\right)=36$

Let E be the event of getting different numbers on dice.

$n\left(E\right)=30=36-6$

$P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)} =\frac{30}{36}=\frac{5}{6}$

(1)

P(happening of the event) + P(not happening of the event) = 1

Sum of probability of happening and non-happening of an event is always 1. So, p + q = 1

(1)

It is given that in a group of 20 people, 5 can not swim.

∴ Number of people who can swim = 20 − 5 = 15

Thus, probability that a person selected at random can swim

$=\frac{15}{12}=\frac{3}{4}$

(1)

When two dice are thrown together, the possible outcomes are:

(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6).

Therefore, the total number of possible outcomes = n(S)=36

Now, the outcomes favourable to event E ‘getting 12 as the product of two numbers obtained’ are
(2, 6), (3, 4), (4, 3), (6, 2).

So, the number of favourable outcomes = n(E)=4

Therefore, the probability of getting 12 as a product of two numbers obtained

$P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{4}{36}=\frac{1}{9}$

(3)

When two coins are tossed, the total possible outcomes are $S=HT,TH,HH,TT\Rightarrow n\left(S\right)=4$

The outcomes favourable to event E‘at least one tail’ are $E=TH,HT,TT\Rightarrow n\left(E\right)=3$

$\therefore \mathrm{The} \mathrm{probability} \mathrm{of} \mathrm{getting} \mathrm{at} \mathrm{least} \mathrm{one} \mathrm{tail} P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{3}{4}$

(1)

The probability of an event which is very unlikely to happen is closest to zero and from the given options 0.0001 is closest to zero.

(3)

When three unbiased coins are tossed together,
Possible outcomes are HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.
The number of all possible outcomes is 8 ⇒ n(S) = 8

(i) The outcomes favourable to the event E ‘exactly two heads’ are HHT, HTH, THH i.e.,  n(E)=3 P(E)= Probability of getting exactly 2 heads =3/8, statement (i) is not true.

(ii) The outcomes favourable to the event F ‘atleast one head’ are HHH, HHT, HTH, THH, HTT, THT, TTH i.e., n(F)=7

P(F)=Probability of getting atleast one head =7/8, statement (ii) is true.

(iii) The outcomes favourable to the event G ‘atmost two tails’ are HHH, HHT, HTH, THH, HTT, THT, TTH i.e., n(G)=7

P(G)=Probability of getting atmost 2 tails =7/8, statement (iii) is true.

(iv) The outcomes favourable to the event H ‘exactly one tail’ are HHT, HTH, THH i.e., n(H)=3

P(H)=Probability of getting exactly one tail =3/8, statement (iv) is true.

So, option (ii), (iii) and (iv) are correct.

(3)

It is given that one card is drawn from a well shuffled deck of 52 cards.
So, the number of possible outcomes = 52

(i) There are 13 cards of diamond in a deck

So, probability of getting a card of diamond = 13/52 = 1/4, statement (i) is not true.

(ii) There is one ace of heart in a deck.
So, probability of getting an ace of heart =1/52, statement (ii) is true.

(iii) There are 39 cards which are not a heart So, probability of not getting a heart = 39/52 = 3/4, statement (iii) is true.

(iv) There are 8 cards which are king or queen. Probability of getting a king or queen = 8/52, 2/13, statement (iv) is not true.

So, options (ii) and (iii) are correct.