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the tree diagram below shows all of the possible outcomes for flipping three coins.

a tree diagram has outcomes (h, h, h), (h, h, t), (h, t, h), (h, t, t), (t, h, h), (t, h, t), (t, t, h), (t, t, t).

what is the probability of one of the coins landing on tails and two of them landing on heads?
1/4
3/8
1/2
3/4

Probability of one of the coin landing on tails and two of them landing on heads is

Step-by-step explanation:

Given:-

Outcomes = (H, H, H), (H, H, T), (H, T, H), (H, T, T), (T, H, H), (T, H, T), (T, T, H), (T, T, T)

To find:- Probability of 1 coin landing on tails and 2 heads=?

Solution:-

Outcomes = (H, H, H), (H, H, T), (H, T, H), (H, T, T), (T, H, H), (T, H, T), (T, T, H), (T, T, T)

    ----------------(1)

Favourable outcomes = (H,H,T),(H,T,H),(T,H,H)

   ------------------(2)

Now using probability formula,

Probability of outcomes = No. of favourable outcomes / Total no. of possible outcomes

Probability of 1 tail and 2 heads =

Probability of 1 tail and 2 heads =

Probability of 1 tail and 2 heads =    ------(from 1 and 2)

Therefore probability of one of the coin landing on tails and two of them landing on heads is

85.560%

step-by-step explanation:

conditional probability is defined as:

p(a|b) = p(a∩b) / p(b)

in other words, the probability that a occurs, given that b has occurred = the probability that both a and b occur / the probability that b occurs.

in this example:

p(turns 64 | turns 44) = p(turns 64 & turns 44) / p(turns 44).

if someone turns 64, they must have already turned 44, so:

p(turns 64 | turns 44) = p(turns 64) / p(turns 44)

on page 42, we see that of the original 100,000, 94944 turn 44, and 81234 turn 64.   therefore:

p(turns 64 | turns 44) = 81234 / 94944

p(turns 64 | turns 44) = 0.85560