15 Questions MCQ Test CSAT Preparation - Test: Probability- 3
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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
Detailed Solution for Test: Probability- 3 - Question 2
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
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
Detailed Solution for Test: Probability- 3 - Question 5
What is the probability of obtaining at least one tail when a coin is tossed five times?
Detailed Solution for Test: Probability- 3 - Question 6
‘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
P( at least one tail ) : 1 - 1/32
= 31/32 will be answer.
If atleast one child in a family with 3 children is a boy then the probability that 2 of the children are boys, is
Detailed Solution for Test: Probability- 3 - Question 10
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
In rolling two dices, find the probability (i) there is at least one ‘6’
Detailed Solution for Test: Probability- 3 - Question 11
(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.
Detailed Solution for Test: Probability- 3 - Question 12
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 1-18. 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.
Detailed Solution for Test: Probability- 3 - Question 13
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