Sure Event : Test Space & Examples

Ans. Arrangement: First we make the example space Ω = {hh, ht , th, tt }

• Second, we compose our occasion. What rudimentary occasions structure the occasion?
• If we figure a little we can see that it is, indeed, a protected occasion because, if we get '" hh " or "tt" with the two coins we get something similar, and in the other two potential causes, " ht " and " the" we get ahead in of the coins.
• That is regardless of the outcome we get, C will consistently occur.

(A) Find the likelihood that the heaviness of an understudy in the class lies in the span of 46 - 50 kg.

(B) Give two occasions in this specific circumstance, one having likelihood 0 and the other having likelihood 1.

Ans. Arrangement: (A) The absolute number of understudies is 38, and the number of understudies with weight in the stretch 46 - 50 kg is 3.

• In this way, P(weight of an understudy is in the span 46 - 50 kg) = 3/38 = 0.07.

(B) For example, consider the occasion that an understudy gauges 30 kg. Since no understudy has this weight, the likelihood of an event of this occasion is 0. Additionally, the likelihood of an understudy weighing more than 30 kg is 38/38 = 1

What is the likelihood of germination of (I) over 40 seeds in a sack? (ii) 49 seeds in a sack? (iii) over 35 seeds in a sack. 

Ans. Arrangement: Total number of sacks is 5.

1. Number of sacks in which more than 40 seeds sprouted out of 50 seeds is 3. P(germination of in excess of 40 seeds in a pack) = 3/5 = 0.6
2.  Number of sacks in which 49 seeds sprouted = 0. P(germination of 49 seeds in a sack) = 0/5 = 0.
3. Number of sacks in which more than 35 seeds sprouted = 5.

Along these lines, the necessary likelihood = 5/5 = 1

a) What is the example space of the investigation.

b)What is the occasion that the picked card is a dark face card.  (3 marks)

Ans. Arrangement :

1. The results in the example space S are 52 cards in the deck.
2. Let E be the occasion that a dark face card is picked. The results in E are Jack, Queen, King of spades or clubs. Emblematically E = {J, Q, K, of spades and clubs} or E = {J♣, Q♣, K♣, J♠, Q♠, K♠}

Ans. Arrangement: Probability of 2 understudies not having the same birthday is P( 2 understudies same birthday )' = 0.992

• The likelihood of 2 understudies having the same birthday P( 2 understudies same birthday) is
• P(E)' = 1 - P(E)
• P(E)=1 - P(E)'
• 1-0.992
• 0.008

Ans. Arrangement: (i) Probability of red balls, P(red) = Favorable result/Total result

• 3/8

(ii) Likelihood of not red balls, P(red)' = 1 - P(red)

• 1 - 3/8
• 5/8

1. The likelihood of a definite occasion is 1
2. The likelihood of an outlandish occasion is 0
3. The likelihood of an occasion lies somewhere in the range of 0 and 1.  (3 marks)

Ans. Arrangement: I. In a definite occasion n(E) =n(S)

• where 'S' is the certain occasion
• Since the number of components in occasion 'E' will be equivalent to the quantity of components in example space
• By the meaning of likelihood
• P(S)=n(E)/n(S)=1
• P(S)=1

2. Since E has no component n(E)=0

• from the meaning of likelihood :
• P(S)=n(E)/n(S)=0/n(S)
• P(S)=0

3. Let 'S' be the example space and 'E' be the occasion

• Then, at that point,
• 0

Ans: It is given that P(E) = 0.01

• P(E) + P (not E) = 1
• 0.01 + P (not E) = 1 ⇒ P (not E) = 1 – 0.01
• 0.99

Thus, probability of 'not E' = 0.99.

Ans. Possible outcome = {1, 2, 3}

∴ P (Getting a number < 4) = 3/6 = ½

Ques. What is the probability of not getting a 3 if you roll a dice. (3 marks)

Ans. There are 6 events that can occur. We can get 1,2,3,4,5,6 when we roll a dice.

• P(getting 3) = 1/6
• Since both the events are complimentary.
• Therefore P(getting 3) + P(not getting 3) = 1
• P(not getting 3) = 1-1/6
• P(not getting 3) = 5/6

Ans. Given to us that,

• P(white balls)= 1/7
• Therefore, P(not white balls) = 1- P(white balls)
• 1-1/7
• 6/7