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Two dice are thrown together .What is the probability that the sum of the number on the two faces is divided by 4 or 6
Clearly n(S)=6*6=36 Let E be the event that the sum of the numbers on the two faces is divided by 4 or 6.Then E={(1,3),(1,5),(2,2),(2,4),(2,6),(3,1),(3,3),(3,5),(4,2),(4,4),(5,1),(5,3),(6,2), (6,6)} n(E)=14. Hence p(e)=n(e)/n(s)=14/36=7/18
Clearly n(S)=6*6=36
Let E be the event that the sum of the numbers on the two faces is divided by 4 or 6.Then
E={(1,3),(1,5),(2,2),(2,4),(2,6),(3,1),(3,3),(3,5),(4,2),(4,4),(5,1),(5,3),(6,2), (6,6)}
n(E)=14.
Hence p(e)=n(e)/n(s)=14/36=7/18
See less(i) an orange flavoured candy?
(ii) a lemon flavoured candy?
(i) an orange flavoured candy?Solution: Zero(ii) a lemon flavoured candy?Solution: 1
(i) an orange flavoured candy?
Solution: Zero
(ii) a lemon flavoured candy?
Solution: 1
See lessWhat is the probability that a number selected from the 101, 102, …, 140 is a multiple of 3 or 4.
1. 19/40
2. 1/2
3. 1/7
4. 8/40
5. 9/70
Correct Answer - Option 2 : 1/2GivenTotal numbers = 40Formula usedP(E) = Number of favourable outcomes/Total number of outcomesCalculationLet E is an event of drawing a card multiple of 3 or 4Number of favourable outcomes = 20 i.e(102, 104, 105, 108, 111, 112, 114, 116, 117, 120, 123, 124, 126, 128,Read more
Correct Answer – Option 2 : 1/2
Given
Total numbers = 40
Formula used
P(E) = Number of favourable outcomes/Total number of outcomes
Calculation
Let E is an event of drawing a card multiple of 3 or 4
Number of favourable outcomes = 20 i.e(102, 104, 105, 108, 111, 112, 114, 116, 117, 120, 123, 124, 126, 128, 129, 132, 135, 136 ,138 and 140)
P(E) = 20/40 = 1/2
See lessRaman well shuffled a pack of 52 cards during his play. He needs a Jack to win a game. Find the probability of getting a jack.
1. 1/13
2. 2/13
3. 1/2
4. 5/13
5. 4/13
Correct Answer - Option 1 : 1/13Given:Cards = 52Formula used:- P(E) = Number of favorable outcomes/Total outcomesCalculation:Total no of favorable outcome is 4 out of 52 cards.⇒ Total no of outcome is 52⇒ Probability of getting “a jack” = 4/52∴ Probability of getting a jack is 1/13
Correct Answer – Option 1 : 1/13
Given:
Cards = 52
Formula used:-
P(E) = Number of favorable outcomes/Total outcomes
Calculation:
Total no of favorable outcome is 4 out of 52 cards.
⇒ Total no of outcome is 52
⇒ Probability of getting “a jack” = 4/52
∴ Probability of getting a jack is 1/13
See lessOne card is drawn from a pack of well shuffle cards. what will be the probability of a diamond card?
1. 1/8
2. 1/4
3. 1/5
4. 1/6
5. 1/2
Correct Answer - Option 2 : 1/4Formula used:nCr = n!/((n-r)! × r!)Calculation:Let S be the sample space then,n(S) = number of way of selecting 1 card out of 52 card⇒ n(S) = 52C1 = 52Let E be the event of getting a diamond cardTotal number of diamond card = 13n(E) = 13C1⇒ 13P(E) = n(E)/n(S) ⇒ P(E) =Read more
Correct Answer – Option 2 : 1/4
Formula used:
nCr = n!/((n-r)! × r!)
Calculation:
Let S be the sample space then,
n(S) = number of way of selecting 1 card out of 52 card
⇒ n(S) = 52C1 = 52
Let E be the event of getting a diamond card
Total number of diamond card = 13
n(E) = 13C1
⇒ 13
P(E) = n(E)/n(S)
⇒ P(E) = 13/52
⇒ 1/4
∴ The probability of a diamond card is 1/4
See lessIf two dice are thrown simultaneously then find the probability that the sum of the numbers coming up on them is 11, given that the number 5 always occurs on the first dice.
1. 1/3
2. 1/36
3. 1/12
4. 1/6
Correct Answer - Option 4 : 1/6Concept:P(A \(\cap\) B) = P(A) x P(B | A) = P(B) x P(A | B) where P(A | B) represents the conditional probability of A given B and P (A | B) represents the conditional probability of B given A.Calculation:Let S be the sample space∴ n(S) = 36Let, A = the event that theRead more
Correct Answer – Option 4 : 1/6
Concept:
P(A \(\cap\) B) = P(A) x P(B | A) = P(B) x P(A | B) where P(A | B) represents the conditional probability of A given B and P (A | B) represents the conditional probability of B given A.
Calculation:
Let S be the sample space
∴ n(S) = 36
Let, A = the event that the sum of the numbers on the two dice is 11.
∴ A = {(5,6), (6,5)}
∴ n(A) = 2
Let, B = the event of the occurrence of 5 on the first dice.
B = {(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)}
∴ n(B) = 6
Now, P(B) = n(B) / n(S) = 6/36 = 1/6.
⇒ A ∩ B = {(5,6)}
∴ n(A∩B) = 1
Now, P(A ∩ B) = n(A ∩ B) / n(S) = 1/36.
⇒ P(A | B) = P(A ∩ B) / P(B) = 1/6
Hence, option 4 is correct.
See lessFor two events, G and Q, it is given that, P(G) = 2/5, P(Q) = 3/8 andP(G|Q) = 2/3 . If G̅ and Q̅ are the complementary events of G and Q, then what is \(\rm P(\frac{\bar G}{\bar Q} )\)equal to?
1. 31/25
2. 8/5
3. 21/40
4. 19/25
Correct Answer - Option 4 : 19/25Concept:P(A/B) = P(A∩B) / P(B)\(\rm \overline{ P(A ∪B)}\)=\(\rm { P(\bar A ∩ \bar B)}\rm= 1-{ P(A ∪ B)}\)P(A ∪ B) = P(A) +P(B) - P(A ∪ B) Calculation: Here, P(G) = 2/5, P(Q) = 3/8 and P(G|Q) = 2/3P(G|Q) = P(G ∩ Q) / P(Q) = 2/3⇒ P(G ∩ Q) = 2/3 × 3/8 = 1/4\(\rm P(\fracRead more
Correct Answer – Option 4 : 19/25
Concept:
P(A/B) = P(A∩B) / P(B)
\(\rm \overline{ P(A ∪B)}\)=\(\rm { P(\bar A ∩ \bar B)}\rm= 1-{ P(A ∪ B)}\)
P(A ∪ B) = P(A) +P(B) – P(A ∪ B)
Calculation:
Here, P(G) = 2/5, P(Q) = 3/8 and P(G|Q) = 2/3
P(G|Q) = P(G ∩ Q) / P(Q) = 2/3
⇒ P(G ∩ Q) = 2/3 × 3/8 = 1/4
\(\rm P(\frac{\bar G}{\bar Q} )=\frac{P(\bar G∩ \bar Q)}{P(\bar Q)}=\frac{\overline {P( G∪ Q) }}{P(\bar Q)}\)
\(\rm =\frac{1-P(G∪ Q)}{P(\bar Q)}\)
P(Q̅) = 1 – P(Q) = 1-3/8 = 5/8
P(G ∪ Q) = P(G) + P(Q) – P(G∩Q)
= 2/5 + 3/8 – 1/4
= 21/40
\(\rm P(\frac{\bar G}{\bar Q} )=\frac{1-\frac {21}{40}}{\frac 58}\)
= 19/40 × 8/5
= 19/25
Hence, option (4) is correct.
See lessA box contains 1 red and 3 identical white balls. Two balls are drawn at random in succession without replacement. Write the sample space for this experiment.
It is given that the box contains 1 red ball and 3 identical white balls. Let us denote the red ball with R and a white ball with W.When two balls are drawn at random in succession without replacement, the sample space is given by S = {RW, WR, WW}
It is given that the box contains 1 red ball and 3 identical white balls. Let us denote the red ball with R and a white ball with W.
See lessWhen two balls are drawn at random in succession without replacement, the sample space is given by S = {RW, WR, WW}
A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be
(i) red ? (ii) white ? (iii) not green?
Solution: Total number of outcomes = 5 + 8 + 4 = 17Number of red marbles = 5Number of white marbles = 8Number of green marbles = 4(i) Red?Solution: Probability of red marbles;P(R) = 5/17(ii) White?Solution: Probability of white marbles;P(W) = 8/17(iii) Not green?Solution: Probability of green marbleRead more
Solution: Total number of outcomes = 5 + 8 + 4 = 17
Number of red marbles = 5
Number of white marbles = 8
Number of green marbles = 4
(i) Red?
Solution: Probability of red marbles;
P(R) = 5/17
(ii) White?
Solution: Probability of white marbles;
P(W) = 8/17
(iii) Not green?
Solution: Probability of green marbles;
P(G) = 4/17
Or, P( not G)
= 1 – 4/17
=13/17
See lessA lot of 4 white and 4 red balls is randomly divided into two halves. What is the probability that there will be 2 red and 2 white balls in each half?
1. 18/35
2. 3/35
3. 1/2
4. None of the above
Correct Answer - Option 1 : 18/35Concept:we know that \(\rm ^nC_r\) = \(\rm \dfrac{n!}{r!(n-r)!}\) Calculations:we are picking the four balls without replacement and that the 4 white and 4 red balls are the only balls in the box and that we choose 2 red balls and 2 white balls.Number of ways to choRead more
Correct Answer – Option 1 : 18/35
Concept:
we know that
\(\rm ^nC_r\) = \(\rm \dfrac{n!}{r!(n-r)!}\)
Calculations:
we are picking the four balls without replacement and that the 4 white and 4 red balls are the only balls in the box and that we choose 2 red balls and 2 white balls.
Number of ways to choose 4 balls out of 8 = \(\rm ^8C_4\) = \(\rm \dfrac{8!}{4!4!}\) = 70
Number of ways to pick 2 red balls out of 4 = \(\rm ^4C_2\) = \(\rm \dfrac{4!}{2!2!}\) = 6
Number of ways to pick 2 white balls out of 4 = \(\rm ^4C_2\) = \(\rm \dfrac{4!}{2!2!}\) = 6
Probability = \(\frac {6 \times 6} {70} = \frac {36} {70} = \frac {18} {35} \)
See less