A fair coin is tossed and an unbiased dice is rolled together. What is the probability of getting a 2 or 4 or 6 along with head?
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NDA (Held On: 21 Apr 2019) Maths Previous Year paper
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- \(\frac{1}{2}\)
- \(\frac{1}{3}\)
- \(\frac{1}{4}\)
- \(\frac{1}{6}\)
Answer (Detailed Solution Below)
Option 3 : \(\frac{1}{4}\)
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Electric charges and coulomb's law (Basic)
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Concept
The sample space when a fair coin is tossed and an unbiased dice is rolled together is as follows:
(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6)
(T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)
Total outcomes = 12
Calculation
Pairs in which we get a 2 or 4 or 6 along with head are as follows:
(H, 2), (H, 4), (H, 6)
Total favourable outcomes = 3
Required probability
\(= \frac{3}{{12}}\)
\(= \frac{1}{{4}}\)
Correct option is (3).
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1) An unbiased dice is thrown. What is the probability that the number is Even
Solution:
Total number of outcomes possible when a die is thrown = 6,
i.e. `S = {1,2,3,4,5,6}`
`:.n(S)=6`
Let `E` = event of getting a number is even
`:.E = {2,4,6}`
`:.n(E) = 3`
`:.P(E)=(n(E))/(n(S))=3/6=1/2`
2) An unbiased dice is thrown. What is the probability that the number is Multiple/Divide by 3Solution:
Total number of outcomes possible when a die is thrown = 6,
i.e. `S = {1,2,3,4,5,6}`
`:.n(S)=6`
Let `E` = event of getting a number divide by 3
`:.E = {3,6}`
`:.n(E) = 2`
`:.P(E)=(n(E))/(n(S))=2/6=1/3`
3) An unbiased dice is thrown. What is the probability that the number is <= 4Solution:
Total number of outcomes possible when a die is thrown = 6,
i.e. `S = {1,2,3,4,5,6}`
`:.n(S)=6`
Let `E` = event of getting a number `<=` 4
`:.E = {1,2,3,4}`
`:.n(E) = 4`
`:.P(E)=(n(E))/(n(S))=4/6=2/3`
4) An unbiased dice is thrown. What is the probability that the number is PrimeSolution:
Total number of outcomes possible when a die is thrown = 6,
i.e. `S = {1,2,3,4,5,6}`
`:.n(S)=6`
Let `E` = event of getting a number is prime
`:.E = {2,3,5}`
`:.n(E) = 3`
`:.P(E)=(n(E))/(n(S))=3/6=1/2`
5) An unbiased dice is thrown. What is the probability that the number is >= 5Solution:
Total number of outcomes possible when a die is thrown = 6,
i.e. `S = {1,2,3,4,5,6}`
`:.n(S)=6`
Let `E` = event of getting a number `>=` 5
`:.E = {5,6}`
`:.n(E) = 2`
`:.P(E)=(n(E))/(n(S))=2/6=1/3`
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- Quant Probability
An unbiased dice is thrown. What is the probability of getting
- (i) an even number
- (ii) a multiple of 3
- (iii) an even number or a multiple of 3
- (iv) an even number and a multiple of 3
Solution
In a single throw of an unbiased dice, you can get any one of the outcomes: 1, 2, 3, 4, 5, or 6.
So, exhaustive number of cases = 6.
(i) An even number is obtained if you obtain any one of 2, 4, 6 as an outcome.
So, favourable number of cases = 3.
Thus, required probability = 3/6 = 1/2
(ii) A multiple of 3 is obtained if you obtain any one of 3, 6 as an outcome.
So, favourable number of cases = 2
Thus, required probability = 2/6 = 1/3
(iii) An even number or a multiple of 3 is obtained in any of the following outcomes 2, 3, 4, 6.
So, favourable number of cases = 4.
Thus, required probability = 4/6 = 2/3
(iv) An even number and a multiple of 3 is obtained if you get 6 as an outcome.
So, favourable number of cases = 1.
Thus, required probability = 1/6
Solution
Difference between biased and unbiased dice:
In a biased die, the outcomes are not equally likely. So that the probability of coming to each face is not equal.
An unbiased dice means that there is an equal probability of occurrence of any of the faces when the dice is rolled.
So that, when we rolled an unbiased dice, the probability of coming to each face is equal.
∴P(1)=P(2)=P(3)=P(45)=P(5)=P(6)=16
Hence, In the biased die, the outcomes are not equally likely and in the unbiased die, outcomes are equally likely.