Probability

Every time a coin is tossed there is a 50 / 50 chances of it coming up heads. There is no rule that says tossing it 100 or 6 times will change this.

go and jump into the well

Each outcome has a probability of 0.05

The answer depends on how many sides the spinner has.

5040

52 cards in a pack.

4 two's, 4 three's, hence 8 chances of drawing one of the desired cards out of the 52.

8/52 = 2/13

With fair regular dice, the answer is 4/36 = 1/9

Suppose the random variable W represents the weight, and assume that W is Normally distributed with the given mean and standard deviation.

Then Prob(140 < W < 220) = Prob[(140 - 145)/31 < Z < (220 - 145)/31]

where Z has a standard Normal distribution.

that is, Prob(-0.1613 < Z < 2.4194) = 1 - [Prob(-0.1613 < Z) + Prob(Z < 2.4194)

= 1 - (0.4359 + 0.0078) = 0.5563

So percentage of women = 100*0.5563 = 56% approx.

In a deck of 52 cards, there are 4 chances to get an ace and 4 chances to get a ten so there are 8 chances to get an ace or a ten

So the probability is 8/52=2/13=0.15

(7!/(4!*3!))*(1/2)^3 * (1/2)^4 = 35*(1/2)^7 which is about 0.2734375

Every time you flip a coin it has a 50% probability that it will land on either heads or tails. You would expect to get heads about half the time and tails about half the time. What actually happens could be different from what is expected. You could get heads every time, or tails every time. Or you could get tails 75% of the time and heads 25% of the time. however, your results appear to be what you would expect, approximately 50% heads and 50% tails. You got 16 heads and 14 tails. Your percentage of heads is 16/30 x 100= 53.33... %. Your percentage of tails is 14/30 x 100 = 46.66... %.

There are 6 values, 1 - 6, on the die. Less than 3 is a 1 or 2 roll. So, probability of less than 3 is 2/6 or 1/3 or 33.33%.

The answer is 1/9 because the possibilities are (1,4), (2,3), (3,2), and (4,1). 4 outcomes out of a total 36 possible: 4/36 = 1/9 = 11.11% chance

$2 to $22,500 depending upon the condition of the coin.

Yes, it is an independent event. However, if you did not replace the card it would not be an independent event, since you don't have an equal chance of pulling any particular card.

The outcome space is TT, TH, HT, HH: 4 outcomes in all.

{One head} = {TH, HT} with prob = 2/4 = 1/2

{No heads} = {TT} with prob 1/4

So E = 2*Pr(one H) - 4*Pr(no H) = 2*(1/2)-4*(1/4) = 1 - 1 = 0

It would depend on the location and crime rate.

Check the local crime statistics for the area.

The probability of an event must be a number in the interval [0, 1].

It is (1/6)2 = 1/36

it could go either way because in games you don't play with it, but in trash some people say it is a free card.

1/48 if there aren't jokers

a biased sample is chosen because it is liked more (looks prettier, more popular etc.) and a random one has no prefeerence hope i helped!!!! :p

It represents the total number of outcomes of a trial divided by the highest common factor of the number of favourable outcomes and the total number of outcomes.