Probability Test - 3

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.

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​

(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.

Total sample space ={B1​B2​,B1​G2​,G1​B2​}
 

Favourable ways ={B1​G2​,G1​B2​}
 

∴ Required probability =2/3.

So, the correct answer is C.​

‘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.

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

(i) Since 2 is the only prime number out of the three numbers, the answer would be 1/3

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.

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.

(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.

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.

1/6 + 1/10 + 1/8 = 47/120

(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

1/⁴C₂ = 1/6.