Here,P(A)=4/9=p/(p+q)
P(odd against event A)=q/p=5/4
We know that:
When one card is drawn from a pack of 52 cards
The numbers of possible outcomes n(s) = 52
We know that there are 26 red cards in the pack of 52 cards
? The numbers of favorable outcomes n(E) = 26
Probability of occurrence of an event: P(E)=Number of favorable outcomes/Numeber of possible outcomes=n(E)/n(S)
? required probability = 26/52 = 1/2.
Here, n(E) =
And, n(S) =
P(S) =
= 7/44
Probability that only one of them is selected = (prob. that brother is selected) × (prob. that sister is not selected) + (Prob. that brother is not selected) × (Prob. that sister is selected)
= =
n(S) = 20
n(Even no) = 10 = n(E)
n(Prime no) = 8 = n(P)
P(E U P) = 10/20 + 8/20 - 1/20 = 17/20
Number of spades in a standard deck of cards=13
Number of aces in a standard deck of cards=4
And,one of the aces is a spade.
So, 13 + 4 - 1 = 16 spades or aces to choose from.
Therfore,probabiltiy of getting a spade or an ace=16/52=4/13
In a family with 2 children there are four possibilities:
1) the first child is a boy and the second child is a boy (bb)
2) the first child is a boy and the second child is a girl (bg)
3) the first child is a girl and the second child is a boy (gb)
4) the first child is a girl and the second child is a girl (gg)
But already given that one child is boy. So we have three possibilities of (bb)(bg)(gb).
n(E)= both are boys=BB=1
n(S)= 3
Required probability P = n(E)/n(S) = 1/3.
The number of exhaustive outcomes is 36.
Let E be the event of getting an even number on one die and an odd number on the other. Let the event of getting either both even or both odd then = 18/36 = 1/2
P(E) = 1 - 1/2 = 1/2.
Number of ways of (selecting at least two couples among five people selected) = (?C? x ?C?)
As remaining person can be any one among three couples left.
Required probability = (?C? x ?C?)/¹?C?
= (10 x 6)/252 = 5/21
In a simultaneous throw of two dice, n(S) = 6 x 6 = 36
Let E = event of getting a doublet = { (1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}
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