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Two coin are tossed simultaneously, find probability that both are tails.
1. 3/4
2. 1/4
3. 3/5
4. 4/7
Correct Answer - Option 2 : 1/4Formula used:P(E) = n(E)/n(S)Calculation: Let S be the sample space Here S = HH, HT, TH, TT⇒ n(S) = 4Let E is event to getting bot are tails = TT⇒ n(E) = 1P(E) = n(E)/n(S)⇒ 1/4∴ Probability of both aRead more
Correct Answer – Option 2 : 1/4
Formula used:
P(E) = n(E)/n(S)
Calculation:
Let S be the sample space
Here S = HH, HT, TH, TT
⇒ n(S) = 4
Let E is event to getting bot are tails = TT
⇒ n(E) = 1
P(E) = n(E)/n(S)
⇒ 1/4
∴ Probability of both are tails is 1/4
See lessA bag has 5 red marbles, 4 green marbles and 3 blue marbles. All marbles are identical in all respects other than colour. A marble is taken out from the bag without looking into it. What is the probability that it is a non-green marble?
1. \(\frac{7}{12}\)
2. \(\frac{5}{12}\)
3. \(\frac{1}{3}\)
4. \(\frac{2}{3}\)
Correct Answer - Option 4 : \(\frac{2}{3}\)Given:The bag has red marbles = 5The bag has green marbles = 4The bag has blue marbles = 3Concept used:The probability formula is used to compute the probability of an event to occur. To recall, the likelihood of an event happening is called probability.ForRead more
Correct Answer – Option 4 : \(\frac{2}{3}\)
Given:
The bag has red marbles = 5
The bag has green marbles = 4
The bag has blue marbles = 3
Concept used:
The probability formula is used to compute the probability of an event to occur. To recall, the likelihood of an event happening is called probability.
Formulae required:
Probability P(A) = the number of favourable outcomes/total number outcomes
Calculations:
The total number of marbles = 5 + 4 + 3
⇒ 12
Non- green marble = not green
The non-green marbles in a bag = total marbles – green marbles
⇒ 12 – 4 = 8
Probability of non-green marble = 8/12
⇒ 2/3
∴ The probability that it is a non-green marble is 2/3
See lessTwo dice are thrown simultaneously. The probability of getting a total of 10 is:
1. 1/4
2. 1/6
3. 1/12
4. 1/9
Correct Answer - Option 3 : 1/12Given:Two dice are thrown simultaneouslyFormula Used:P(E) = n(E)/n(S)Where,n(E) = Number of favourable eventsn(S) = Number of possible outcomesCalculation:When two dice throw simultaneously, then total possible outcomes(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)(2,Read more
Correct Answer – Option 3 : 1/12
Given:
Two dice are thrown simultaneously
Formula Used:
P(E) = n(E)/n(S)
Where,
n(E) = Number of favourable events
n(S) = Number of possible outcomes
Calculation:
When two dice throw simultaneously, then total possible outcomes
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2),…………………………., (2, 6)
……………………………………………………
……………………………………………………
(6, 1), (6, 2), …………………………..(6, 6)
Number of possible outcomes n(S) = 6 × 6
⇒ 36
Number of favourable events n(E) = {(4,6), (6,4), (5,5)} = 3
⇒ n(E) = 3
Required probability of getting a total of 10
P(E) = 3/36
⇒ 1/12
∴ The probability of getting a total of 10 is 1/12.
See lessA die is thrown what is the probability of a getting 5 up?
1. 1/13
2. 3/7
3. 2/3
4. 1/6
Correct Answer - Option 4 : 1/6Formula used:P(E) = n(E)/n(S)Where, n(S) → sample space, n(E) → favourable eventCalculation:Let S be the sample spaceS = 1, 2, 3, 4, 5, 6N(S) = 6Let E be the event of getting 5⇒ n(E) = 1P(E) = n(E)/n(S)⇒ 1/6∴ Probability of a getting 5 up is 1/6
Correct Answer – Option 4 : 1/6
Formula used:
P(E) = n(E)/n(S)
Where, n(S) → sample space, n(E) → favourable event
Calculation:
Let S be the sample space
S = 1, 2, 3, 4, 5, 6
N(S) = 6
Let E be the event of getting 5
⇒ n(E) = 1
P(E) = n(E)/n(S)
⇒ 1/6
∴ Probability of a getting 5 up is 1/6
See lessGiven below are 2 statements:
Statement I: The probability of a sure event is 1.
Statement II: The probability of an impossible event is -1.
In light of the above statements, choose the correct option given below.
1. Both statements I and statement II are true.
2. Both statements I and statement II are false.
3. Statements I is correct but statement II is false.
4. Statement I is false but Statements II is correct.
Correct Answer - Option 3 : Statements I is correct but statement II is false.Concept :Sure event: A sure event is an event, which always happens. The probability of a sure event is always 1.Impossible event: An impossible event is an event that cannot happen. The probability of an Impossible event Read more
Correct Answer – Option 3 : Statements I is correct but statement II is false.
Concept :
Sure event: A sure event is an event, which always happens. The probability of a sure event is always 1.
Impossible event: An impossible event is an event that cannot happen. The probability of an Impossible event is always 0.
Calculation:
Using the above Concepts we can say,
Statements I is correct
Statements II is incorrect.
The correct option is 3 i.e. Statements I is correct but statement II is incorrect.
See lessTwo dice are thrown simultaneously. What is the probability of getting the same number on both the dice?
1. 1/6
2. 1/4
3. 1/3
4. 1/9
Correct Answer - Option 1 : 1/6Given:Two dice are thrown simultaneouslyCalculation:When two dice are thrown simultaneously, Number of possible outcomes are 36.If getting the same number on both dice is taken as event,Then for the 1st event, ⇒ Number of outcomes are 6.For 2nd event,⇒ Number of outcomRead more
Correct Answer – Option 1 : 1/6
Given:
Two dice are thrown simultaneously
Calculation:
When two dice are thrown simultaneously,
Number of possible outcomes are 36.
If getting the same number on both dice is taken as event,
Then for the 1st event,
⇒ Number of outcomes are 6.
For 2nd event,
⇒ Number of outcomes/total number of possible outcomes
⇒ 6/(6 × 6)
⇒ 1/6
∴ The probability of getting the same number of both the dice is 1/6.
See lessFor any two events A and B, the probability that at least one of them occur is 0.6. If A and B occur simultaneously with a probability 0.3, then P(A’) + P(B’) is
1. 0.9
2. 1.15
3. 1.1
4. 1.0
Correct Answer - Option 3 : 1.1Concept:we know that,\(\rm P(A) +P(B) = P(A \cup B) + P(A \cap B)\)P(A') = 1 - P(A) Calculations:Given, For any two events A and B, the probability that at least one of them occur is 0.6⇒ \(\rm P(A \cup B) = 0.6\)and A and B occur simultaneously with a probability 0.3⇒Read more
Correct Answer – Option 3 : 1.1
Concept:
we know that,
\(\rm P(A) +P(B) = P(A \cup B) + P(A \cap B)\)
P(A’) = 1 – P(A)
Calculations:
Given, For any two events A and B, the probability that at least one of them occur is 0.6
⇒ \(\rm P(A \cup B) = 0.6\)
and A and B occur simultaneously with a probability 0.3
⇒ \(\rm P(A \cap B) = 0.3\)
we know that,
\(\rm P(A) +P(B) = P(A \cup B) + P(A \cap B)\)
⇒ \(\rm P(A) +P(B) = 0.6 + 0.3 = 0.9\)
Also, we know that
P(A’) + P(B’) = 1 – P(A) + 1 – P(B)
⇒ P(A’) + P(B’) = 2 – 0.9
⇒ P(A’) + P(B’) = 1.1
Hence, For any two events A and B, the probability that at least one of them occur is 0.6. If A and B occur simultaneously with a probability 0.3, then P(A’) + P(B’) is 1.1
See lessIf a coin is tossed five times then what is the probability that you observe at least one head?
(a) 1/32
(b) 15/32
(c) 23/32
(d) 31/32
Correct option d (31/32)Explanation:Consider solving this using complement. Probability of getting no head = P(all tails) = 1/32 P(at least one head) = 1 – P(all tails) = 1 – 1/32 = 31/32.
Correct option d (31/32)
Explanation:
Consider solving this using complement.
Probability of getting no head = P(all tails) = 1/32
P(at least one head) = 1 – P(all tails) = 1 – 1/32 = 31/32.
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The Rifle shooter Ajay and Bharat take aim at a target. If the probability of Ajay hitting the target is 70% and the probability of B missing the target is 80%. Find the Probability that both Ajay and Bharat hit the target?
1. 24%
2. 20%
3. 14%
4. 7%
Correct Answer - Option 3 : 14%Given:The probability of shooter Ajay hitting the target is 70%Probability of shooter Bharat Missing the target is 80%Formula Used:Probability = number of favourable outcomes /total number of outcome Calculation:The probability of shooter Ajay hitting the target is 70%Read more
Correct Answer – Option 3 : 14%
Given:
The probability of shooter Ajay hitting the target is 70%
Probability of shooter Bharat Missing the target is 80%
Formula Used:
Probability = number of favourable outcomes /total number of outcome
Calculation:
The probability of shooter Ajay hitting the target is 70% = 7/10
⇒ The probability of shooter Ajay missing the target is 30% = 3/10
Probability of shooter Barath Missing the target is 80% = 8/10
⇒ The probability of shooter Barath Hitting the target is 20% = 2/10
Probability of hitting the target will be = 7/10 × 2/10 = 14%
∴The Probability that both Ajay and Bharat hitting the target will be 14%
See lessTwo fair dice are rolled. What is the probability that the total score is a prime number?
1. \(\dfrac{1}{6}\)
2. \(\dfrac{5}{12}\)
3. \(\dfrac{1}{2}\)
4. \(\dfrac{7}{9}\)
Correct Answer - Option 2 : \(\dfrac{5}{12}\)Concept:The probability of the occurrence of an event A out of a total possible outcomes N, is given by: \(P(A) = \rm \dfrac{n(A)}{N}\), where n(A) is the number of ways in which the event A can occur. Calculation:The total number of distinct possible outRead more
Correct Answer – Option 2 : \(\dfrac{5}{12}\)
Concept:
The probability of the occurrence of an event A out of a total possible outcomes N, is given by: \(P(A) = \rm \dfrac{n(A)}{N}\), where n(A) is the number of ways in which the event A can occur.
Calculation:
The total number of distinct possible outcomes (N), when rolling two dice, are: N = 6 × 6 = 36.
The sum of these pairs of outcomes can be a number from 1 + 1 = 2 to 6 + 6 = 12.
The prime numbers from 2 to 12 are (2, 3, 5, 7, 11). The different possibilities to give each of these sums are given below:
Sum = 2:
(1, 1) = 1 possibility.
Sum = 3:
(1, 2), (2, 1) = 2 possibilities.
Sum = 5:
(1, 4), (4, 1), (2, 3), (3, 2) = 4 possibilities.
Sum = 7:
(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3) = 6 possibilities.
Sum = 11:
(5, 6), (6, 5) = 2 possibilities.
Total Number of possibilities for the desired event to occur is: n(A) = 1 + 2 + 4 + 6 + 2 = 15.
The required probability is therefore: \(P = \rm \dfrac{n(A)}{N}=\dfrac{15}{36}=\dfrac{5}{12}\).
The following table gives the number of ways in which a given sum can be obtained when rolling a pair of dice:
The number 7 has the maximum number of possibilities.
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