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A speaks the truth 3 out of 4 times, and B5 out of 6 times : what is the probability that they will contradict each other in stating the same fact?
Probability that A speak truth,P(A)=3/4
Probability that B speak truth,P(B)=5/6
Probability that A and B lie is
P(A′) lies=1/4 P(B′) lies=1/6
Required probability=P(A)×P(B′)+P(A′)×P(B)
=3/4×1/6+1/4×5/6
⇒8/24⇒1/3
So, the correct option is B.
Three groups of children contain respectively 3 girls & 1 boy; 2 girls & 2 boys; 1 girl & 3 boys. One child is selected at random from each group.The chance that the three selected children consists of one girl and 2 boys is
Let:
Group A: 3 girls, 1 boy
Group B: 2 girls, 2 boys
Group C: 1 girl, 3 boys
The combination of 1 girl and 2 boys can be selected in:
(i) A  girl, B  boy, C  boy: 3/4×2/4×3/4=18/64
(ii) A  boy, B  girl, C  boy: 1/4*2/4*3/4 =6/64
(iii) A  boy, B  boy, C  girl: 1/4*2/4*1/4= 2/64.
Hence, required probability = 18/64+6/64+ 2/64 =26/64=13/32
So, the correct answer is c
From a pack of 52 cards, two are drawn at random. Find the chance that one is a knave and the other a queen.
Three coins are tossed. What is the probability of getting (i) neither 3 Heads nor 3 Tails?
(i) Probability of 3 heads = 1/8 Also, Probability of 3 tails =1/8.
Required probability = 1 (1/8 + 1/8) = 6/8 = 3/4.
Ram is visiting a friend. Ram knows that his friend has 2 children and 1 of them is a boy. Assuming that a child is equally likely to be a boy or a girl, then the probability that the other child is a girl, is
Total sample space ={B1B2,B1G2,G1B2}
Favourable ways ={B1G2,G1B2}
∴ Required probability =2/3.
So, the correct answer is C.
What is the probability of obtaining at least one tail when a coin is tossed five times?
‘At least one tail’ means that there can be one, two or three or four or five tails.
The only option that is not included is five heads.
The sum of all probabilities is always 1
P(at least one tail) = 1  P(no tail)
P(no tail) = P(all heads) = P(H, H ,H, H ,H)
In any throw the probability of tail or head is ½
P(H, H ,H, H ,H) = 1/2 * 1/2 * 1/2 * 1/2 * 1/2 = (1/2)^{5}
= 1/32
P( at least one tail ) : 1  1/32
= 31/32 will be answer.
Two fair dice are thrown. Find the probability of getting
(i) sum of numbers divisible by 2 or 4.
Positive outcomes are 2(1 way), 4(3 ways), 6(5 ways) 8(5 ways), 10 (3 ways), 12 (lway). Thus, 18/36= 1/2
Three cards numbered 2, 4 and 8 are put into a box. If a card is drawn at random, what is the probability that the card drawn is
(i) a prime number?
(i) Since 2 is the only prime number out of the three numbers, the answer would be 1/3
Two fair dice are thrown. Find the probability of getting
(i) a prime number less than 8.
Positive outcomes are 2(1 way), 3(2 ways), 5(4 ways), 7(6 ways). Total of 13 positive outcomes out of 36.
Thus, 13/36.
If atleast one child in a family with 3 children is a boy then the probability that 2 of the children are boys, is
total no of ways are 8 but given that at least 1 boy is present so 1 case is excluded in which all three are girls
so total no ways in which at least one boy is present is 7
total no ways in which 1 girl and 2 boys are present are 3 ways
so the probability is 3/7
So, the correct option is A.
In rolling two dices, find the probability (i) there is at least one ‘6’
(i) With a six on the first dice, there are 6 possibilities of outcomes that can appear on the other dice (viz. 6 & 1, 6 & 2, 6 & 3, 6 & 4, 6 & 5 and 6 & 6). At the same time with 6 on the second dice there are 5 more possibilities for outcomes on the first dice: (1 & 6 , 2 & 6 , 3 & 6, 4 & 6, 5 & 6)
Also, the total outcomes are 36. Hence, the required probability is 11/36.
Amit throws three dice in a special game of Ludo. If it is known that he needs 15 or higher in this throw to win then find the chance of his winning the game.
Event definition is: 15 or 16 or 17 or 18. 15 can be got as: 5 and 5 and 5 (one way) Or 6 and 5 and 4 (Six ways) Or 6 and 6 and 3 (3 ways) Total 10 ways. 16 can be got as: 6 and 6 and 4 (3 ways) Or 6 and 5 and 5 (3 ways) Total 6 ways. 17 has 3 ways and 18 has 1 way of appearing. Thus, the required probability is:(10 + 6 + 3 + l)/216 = 20/216 = 5/54.
In a horse race there were 18 horses numbered 118. The probability that horse 1 would win is 1/6, that 2 would win is 1/10 and that 3 would win is 1/8.
Assuming that a tie is impossible, find the chance that one of the three will win.
1/6 + 1/10 + 1/8 = 47/120
Two fair dice are thrown. What is the probability of (i) throwing a double?
(i) There are six doubles (1, 1; 2 & 2; 3 & 3; 4 & 4; 5 & 5; 6 & 6) out of a total of 36 outcomes
6/36 = 1/6
Two letters are randomly chosen from the word LIME. Find the probability that the letters are L and M.
1/^{4}C_{2} = 1/6.
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