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0.2

If A and B are two events associated with a random experiment such that p(A) = 0.3, p(B) = 0.2 and p(A ∩ B) = 0.1, then the value of p(A – B) is 0.2.

No. of two digits number = 90 

⇒ n(S) = 90. 

Let ‘A’ be the event of getting a number divisible by 15. 

⇒ A = {15, 30, 45, 60, 75, 90} 

⇒ n(A) = 6 

∴ p(A) = \(\frac{n(A)}{n(S)}\) = \(\frac{6}{90}\) = \(\frac{1}{15}\)

∴ The probability of getting a number divisible by 15 = \(\frac{1}{15}\) .

Given:

⇒ Husband tells lies in 30% of cases

⇒ Wife tells lies in 35% of cases

Now,

⇒ P(N_H) = P(Husband telling lies) = 0.30

⇒ P(T_H) = P(Husband telling truth) = 1 - 0.30

⇒ P(T_H) = 0.70

⇒ P(N_W) = P(Wife telling lies) = 0.35

⇒ P(T_W) = P(Wife telling truth) = 1 - 0.35

⇒ P(T_W) = 0.65

We need to find the probability for case in which husband and wife contradict each other for narrating a fact.

This happens only when husband not telling truth while Wife is telling truth and vice-versa.

⇒ P(C_HW) = P(Husband and Wife contradict each other)

⇒ P(C_HW) = P(Husband tells truth and Wife doesn’t) + P(Wife tells truth and Husband doesn’t)

Since the speaking of husband and Wife are independent events their probabilities will multiply each other.

⇒ P(C_HW ) = (P(T_H)P(N_W)) + (P(N_H)P(T_W))

⇒ P(C_HW) = (0.70 × 0.35) + (0.30 × 0.65)

⇒ P(C_HW) = 0.245 + 0.195

⇒ P(C_HW) = 0.44

∴ The required probability is 0.44.

Number days for which the concentration of sulphur dioxide was in the interval of 0.12 − 0.16 = 2

Total number of days = 30

Hence, required probability, P = 2/30 = 1/15

Let the shooter fire n times. Obviously, n fires are n Bernoulli trials. In each trial, p = probability of hitting the target = 3/4 and q = probability of not hitting the target = 1/4 .

Therefore P(X = x) = ⁿC_xq^{n - x}p^x, x = 0, 1, 2, ...., n

= ⁿC_x(1/4)^{n - x}(3/4)^x = ⁿC_x(3^x/4ⁿ)

Now, given that, P(hitting the target at least once) > 0.99 

i.e. P(x ≥ 1) > 0.99 

Therefore, 1 – P (x = 0) > 0.99

⇒1 - ⁿC₀(1/4ⁿ) > 0.99 ⇒ ⁿC₀(1/4ⁿ) < 0.01

⇒ 1/4ⁿ < 0.01 ⇒ 4ⁿ > 1/0.01 = 100

The minimum value of n to satisfy the inequality (1) is 4. 

Thus, the shooter must fire 4 times.

Sample Space = 36 

Out of 36 possible events, only one event is favourable = (6, 6) 

Probability = \(\frac{1}{36}\)

The number of aces is a random variable. Let it be denoted by X. Clearly, X can take the values 0, 1, or 2. 

Now, since the draws are done with replacement, therefore, the two draws form independent experiments. 

Therefore, P(X = 0) = P(non-ace and non-ace) = P(non-ace) × P(non-ace) = 48/52 x 48/52 = 144/169

P(X = 1) = P(ace and non-ace or non-ace and ace)

= P(ace and non-ace) + P(non-ace and ace)

= P(ace). P(non-ace) + P (non-ace) . P(ace)

= 4/52 x 48/52 + 48/52 x 4/52 = 24/169

and P(X = 2) = P (ace and ace) = 4/52 x 4/52 = 1/169

Thus, the required probability distribution is

First six positive integers are 1, 2, 3, 4, 5, 6 

If two numbers are selected at random from above six numbers then sample space S is given by 

S = {(1, 2)(1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 3), (2, 4),(2, 5), (2, 6), (3, 1),(3, 2), (3, 4),(3, 5), (3, 6),(4, 1), (4, 2), (4, 3), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}

n(s) = 30. 

Here, X is random variable, which may have value 2, 3, 4, 5 or 6. 

Therefore, required probability distribution is given as 

P(X = 2) = Probability of event getting (1, 2), (2, 1) = 2/30 

P(X = 3) = Probability of event getting (1, 3), (2, 3), (3, 1), (3, 2) = 4/30 

P(X = 4) = Probability of event getting (1, 4), (2, 4), (3, 4), (4, 1), (4, 2), (4, 3) = 6/30 

P(X = 5) = Probability of event getting (1, 5), (2, 5), (3, 5), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4) = 8/30 

P(X = 6) = Probability of event getting (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5) = 10/30 

It is represented in tabular form as

Required mean = E(x) = ∑p_ix_i = 2 x 2/30 + 3 x 4/30 + 4 x 6/30 + 5 x 8/30 + 6 x 10/30

= (4 + 12 + 24 + 40 + 60)/30 = 140/30 = 14/3 = 4(2/3)

Total number of primary outcomes of the random experiment of throwing a balanced coin thrice is n = 2³ = 8.

• Head (H) obtained in all three trials, i.e., outcome HHH is obtained.
• Head (H) obtained in first two trials and Tail (T) obtained, in third trial, i.e., outcome HHT is obtained.
• Head (H) obtained in first trial, Tail (T) obtained in second trial and Head (H) obtained in third trial, i.e., outcome HTH is obtained.
• Tail (T) obtained in first trial and Head (H) obtained in rest of the two trials, i.e., outcome THH is obtained.
• Head (H) obtained in first trial and Tail (T) obtained in rest of the two trials, i.e., outcome HTT is obtained.
• Tail (T) obtained in first trial, Head (H) obtained in second trial and Tail (T) obtained in third trial, i.e., outcome THT is obtained.
• Tail (T) obtained in first two trials and Head (H) obtained in third trial, i.e., outcome TTH is obtained.
• Tail (T) obtained in all three trials, i.e., outcome TTT is obtained.

Hence, the sample space for the random experiment of throwing a balanced coin is expressed as follows:
U = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

(a) The outcomes in the sample space S are 52 cards in the deck.

(b) Let E be the event that a black face card is chosen. The outcomes in E are Jack, Queen, King or spades or clubs. Symbolically

E = {J, Q, K, of spades and clubs} or E = {J♣, Q♣, K♣, J♠, Q♠, K♠}

(c) \(\frac{(^{2n}C_n)^2}{^{4n}C_{2n}}\)

Total number of boys and girls = 2n + 2n = 4n 

Since, there are two equal batches, each batch has 2n members 

∴ Let S (Sample space) : Selecting one batch out of 2 

⇒ S : Selecting 2n members out of 4n members. 

⇒ n(S) = ⁴ⁿC_2n 

If each batch has to have equal number of boys and girls, each batch should have n boys and n girls. 

Let E : Event that each batch has ‘n’ boys and ‘n’ girls

⇒ n(E) = ²ⁿC_n × ²ⁿC_n = (²ⁿC_n)²

∴ Required probability = \(\frac{n(E)}{n(S)}\) = \(\frac{(^{2n}C_n)^2}{^{4n}C_{2n}}\)

Total number of ways of choosing a group is:  ⁴ⁿC_n

The number of ways in which each group contains equal number of boys and girls is: 

(²ⁿC_n)(²ⁿC_n)

∴ Required probability is (²ⁿC_n)² / ⁴ⁿC_n 

optoin c is correct

HOPE IT HELPS

(i) Even numbers occur is (2,2) (2,4) (2,6) (4,2) (4,4)

(4,6) (6,2) (6,4) (6,6)

P (number of each die is even) = 9/36 = 1/4

(ii) Sum of numbers is 5 in (1, 4) (2,3) (3,2) (4,1) 

P (sum of numbers appearing on two dice is 5

= 4/36 = 1/9

Since total number in English alphabet is 26 in which 5 vowels and 21 consonants.

(i) 5/26

(ii) 21/26

When a card is drawn from a well shuffled deck of 52 cards, the number of possible outcomes is 52. 

(i) Let A : ‘the card drawn is a diamond’ 

Clearly the number of elements in set A is 13. 

∵ n(S) = 52 and n(A) = 13

∴ P(A) = n(A)/n(S) = 13/52 = 1/4

(ii) Let B: ‘card drawn is an ace’. B’: ‘card drawn is not an ace’ we have, n(13) = 4,n(S)= 52

∴ P(B) = n(B)/n(S) = 4/52 = 1/13

Required probability =P(B’)

We have P(B') = 1 - P(B) = 1 - 1/13 = 12/13

(iii) Let C : ‘card drawn is black card’. 

∴ Number of elements in set C = 26 i.e., n(C) =26

∴ P(C) = n(C)/n(S) = 26/52 = 1/2

(iv) Let A: ‘card drawn is a diamond’ 

∴ A’: card drawn is not  a diamond’ 

We have n(A) = 13, n(S) = 52

∴ P(A) = 13/52 = 1/4

we have P(A') = 1 - P(A) = 1 - 1/4 = 3/4

(v) We have p(C) = 1/2  (from (iii))

∴ P(C') = 1 - P(C) = 1 - 1/2 = 1/2

Let 5 red pens be R₁ , R₂ , R₃ , R₄ , R₅ . 8 blue pens be B₁ , B₂ , B₃ , B₄ , B₅ , B₆ , B₇ , B₈ . and 3 green pens be G₁ , G₂ , G₃ . 

Sample space, 

S = {R₁ , R₂ , R₃ , R₄ , R₅, B₁ , B₂ , B₃ , B₄ , B₅ , B₆ , B₇ , B₈, G₁ , G₂ , G₃ } 

∴ n(S) = 16 

Let A be the event that Rutuja picks a blue pen. 

∴ A = {B₁ , B₂ , B₃ , B₄ , B₅ , B₆ , B₇ , B₈}

∴ n(A) = 8

∴ P(A) = \(\frac{n(A)}{n(S)}\)= 8/16

∴ P(A) = 1/2 

∴ The probability that Rutuja picks a blue pen is 1/2 .

(i) Event: Any subset E of a sample space S is called an event. 

(ii) Impossible even: The empty set φ describes an events. Then φ is called an impossible event. 

(iii) Sure event: The sample space S describes an event. Then S, sample space is called the sure event. 

(iv) Simple event: If an event E has only one sample point of a sample space, it is called a simple event. 

(v) Compound event: If an event has more than one sample point, it is called a compound event.

In a leap year, there are 366 days i.e., 52 weeks and 2 days.

In 52 weeks, there are 52 Tuesdays. 

Therefore, the probability that the leap year will contain 53 Tuesdays is equal to the probability that the remaining 2 days will be Tuesdays. 

The remaining 2 days can be

 Monday and Tuesday

 Tuesday and Wednesday 

Wednesday and Thursday 

Thursday and Friday 

Friday and Saturday 

Saturday and Sunday 

Sunday and Monday 

Total number of cases = 7 

Favourable cases = 2

Probability that a leap year will have 53 Tuesdays = 2/7

The box contains 2 red balls and 3 black balls. Let us denote the 2 red balls as R₁, R₂ and the 3 black balls as B₁, B₂, and B₃.
The sample space of this experiment is given by
S = {TR₁, TR₂, TB₁, TB₂, TB₃, H₁, H₂, H₃, H₄, H₅, H₆}

Required sample space = S = {TR1, TR2, TB1, TB2, TB3, H1, H2, H3, H4, H5, H6} 

Here R₁,R₂ are two red balls and B₁,B₂,B₃ are three black balls.

(i) (5,5),(5,6),(6,5),(6,6)

P(a number greater than 4 on each die) = 4/36

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

P(a doubtlet (the same number appearing on each of the die)) = 6/36

If 1 appears on the first drawn slip, then the possibilities that the number appears on the second drawn slip are 2, 3, or 4. Similarly, if 2 appears on the first drawn slip, then the possibilities that the number appears on the second drawn slip are 1, 3, or 4. The same holds true for the remaining numbers too.
Thus, the sample space of this experiment is given by
S = {(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3)}

(i) P(the number 5) = 1/30

(ii) 1,3,5,7,9,11,13,15,17,19,21,23,25,27,29

P(an odd number) = 15/30

(iii) 2,3,5,7,11,13,17,19,23,29

P(a prime number) = 10/30

Given: There are 4 slips in the box and mixed thoroughly. 

To Find: Describe the given events . 

Explanation: Here, Four slips of paper 1, 2, 3, and 4 are put in a box. 

If Two slips are drawn from it one after the other without replacement. 

Then The sample space for the experiment is: 

S = {(1, 2)(1, 3), (1, 4)(2, 1)(2, 3)(2, 4)(3, 1)(3, 2)(3, 4)(4, 1)(4, 2)(4, 3)} 

(i) A = number on the first slip is larger than the one on the second slip, 

So, The sample space for A is: 

{(2, 1)(3, 1)(3, 2)(4, 1)(4, 2)(4, 3)} 

(ii) B = number on the second slip is greater than 2 

So, The sample space for B is 

{(1, 3)(2, 3)(1, 4)(2, 4)(3, 4)(4, 3)} 

(iii) C = sum of the numbers on the two slips in 6 or 7 

The sample space for C is 

{(2, 4)(3, 4)(4, 2)(4, 3)} 

(iv) D = number on the second slip is two times the number on the first slip The sample space if: 

{(1, 2)(2, 4)} 

Now, 

We can see, AՌD = Փ.

Therefore, A and D are mutually exclusive events 

Hence, A and D are mutually exclusive events.

(c) \(\frac{3}{16}\)

Total number of determinants that can be formed using 0 and 1 = 16 (4 × 4)

The positive determinants are \(\begin{bmatrix}1&0\\[0.3em]1&1\end{bmatrix}\),\(\begin{bmatrix}1&0\\[0.3em]0&1\end{bmatrix}\),\(\begin{bmatrix}1&1\\[0.3em]0&1\end{bmatrix}\), i.e, 3 in number.

∴ Required probability = \(\frac{3}{16}.\)

When two dice are thrown, the sample space is 

S = {(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)} 

∴ n(S) = 36 

(a) Let event A: Sum of the numbers on uppermost face is 5. 

∴ A = {(1, 4), (2, 3), (3, 2), (4, 1)} 

∴ n(A) = 4

∴ P(A) = \(\frac {n(A)}{n(S)} = \frac {4}{36} = \frac {1}{9}\)

(b) Let event B: Sum of the numbers on uppermost face is at least 8 (i.e., 8 or more than 8) 

∴ B = {(2, 6), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6), (5, 3), (5, 4), (5, 5), (5, 6), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

∴ n(B) = 15

 ∴ P(B) = \(\frac {n(B)}{n(S)} = \frac {15}{36} = \frac {5}{12}\)

(c) Let event C: First throw gives a multiple of 2 and second throw gives a multiple of 3. 

∴ C = {(2, 3), (2, 6), (4, 3), (4, 6), (6, 3), (6, 6)} 

∴ n(C) = 6

  ∴ P(C) = \(\frac {n(C)}{n(S)} = \frac {6}{36} = \frac {1}{6}\)

(d) Let event D: The product of the numbers on uppermost face is 12. 

∴ D = {(2, 6), (3, 4), (4, 3), (6, 2)} 

∴ n(D) = 4

   ∴ P(C) = \(\frac {n(D)}{n(S)} = \frac {4}{36} = \frac {1}{9}\)

Correct option   (c)   9/25

Explanation:

Let two non-negative integers be x & y such that x = 5a + α and y = 5b  + β where 0 ≤ α, ≤ β

New, x² + y² = 25(a² + b² ) + 10(aα + bβ)+ α² + β²

α² + β² should be divisible by 5

(α ,β) can have ordered pairs as

{(0,0), (1,2), (2,1), (1,3), (3,1), (2,4), (4,2), (3,4), (4,3)} = 9 ways

 required probability is 9/25

Correct option  (b) 7/12

Explanation:

Favourable ways is 21.

 required probability is 21/36 = 7/12